Relativistic KE for omnidirectional feats

[Floxy178]

As far as I know, we don't have a method for calculating omnidirectional feats when speed is relativistic. I've seen it being calculated as being same fraction as classical KE formula (like calculating relativistic KE and dividing it by 2 or 6), but it isn't correct, as KE isn't quadratic with speed here, so distribution will be different from how those derived 1/4 or 1/12 treat it.

Difference becomes larger as speed is closer to c (method mentioned above gets higher results than actual), as for these omnidirectional feats, rise of KE isn't that big with speed approaching c compared to normal rel KE because most of its speed is significantly lower than c.

Spoiler: Case 1

Expanding from center (1/4 version)

Speed at distance r from center:
v(r) = (r/R)V

Gamma factor at distance r:
γ(r) = 1/√(1 - v(r)²/c²) = 1/√(1 - V²r²/(R²c²))

I'll use all parts that are same distance from center (AKA with same speed) together:

KE = ∫₀ᴿ (γ(r) - 1)c² ϱ 2πr H dr
KE = 2πϱHc² ∫₀ᴿ (γ(r) - 1)r dr

Simply, taking 2πrϱH's mass (which all parts have same speed) and using speed for its r to find relativistic KE of it. And applying this for all r from r=0 to r=R.

KE =
2πϱHc² ∫₀ᴿ (1/√(1 - V²r²/(R²c²)) - 1)r dr

Integration part gives:
(R^2 * c^2 / V^2) * (1 - sqrt(1 - V^2/c^2)) - R^2/2

Then you will multiply it by 2πϱHc².

Spoiler: Case 2

Dispersing them from center (1/12 version)

Same as Case 1, just here:
v(r) = V(1-r/R)

So formula becomes:
KE =
2πρHc² ∫₀ᴿ (1/√(1 - (V²/c²)(1 - r/R)²) - 1) r dr

Integration gives:
R^2 * ((c * arcsin(V/c)) / V + (c^2 * (1 - sqrt(1 - V^2/c^2))) / V^2 - 1/2 )

Then you will multiply by 2πϱHc².

I'll also add versions for sphere shape (omnidirectional 3D wise). Almost the same, just for taking parts that have same speed we use 4πr²ρ instead of 2πrHρ. Even if not clouds I'm sure there'll be some feats like that with air.

Spoiler: Case 1 for Sphere

Expanding case

if v<<c use:
Newtonian = ∫₀ᴿ 0.5 * 4 * π * r^2 * ρ * (v * r / R)^2 dr = 2 * π * ρ * v^2 / R^2 * ∫₀ᴿ r^4 dr = 2 * π * ρ * v^2 / R^2 * R^5/5 = 3/10 mv^2
(since ρ = 3m/4πR^3)

Relativistic KE =
4πρc² ∫₀ᴿ (1/√(1 - V²r²/(c²R²)) - 1) r² dr

Integration part:
(R^3 * c^3 / V^3) * [ (1/2)arcsin(V/c) - (V/(2c))√(1 - V^2/c^2) - (V^3)/(3c^3) ]

Then multiply by 4πρc².

Spoiler: Case 2 for Sphere

Dispersing from center case

if v<<c use:
Newtonian = ∫₀ᴿ 0.5 * 4 * π * r^2 * ρ * (v * (1 - r / R))^2 dr = 2 * π * ρ * v^2 * ∫₀ᴿ r^2(1 - r / R)^2 dr = 2 * π * ρ * v^2 * R^3/30 = 1/20 mv^2

Relativistic KE =
4πρc² ∫₀ᴿ (1/√(1 - (V²/c²)(1 - r/R)²) - 1) r² dr

Integration part:
R^3 * ( (c / V) * [ arcsin(V / c) - 2 * (c / V) * (1 - sqrt(1 - V^2 / c^2)) + (c^2 / V^2) * (0.5 * arcsin(V / c) - (V / (2 * c)) * sqrt(1 - V^2 / c^2)) ] - 1/3)

Then multiply by 4πρc².

Alternatively, I hope maybe someone could make calculator for these.

Edit:

Accounting for length contraction was proposed by DT so I'll write here new versions. All of them have just additional γ(r) in formula. Everything else is the same as above.

Case 1:
2πϱHc² ∫₀ᴿ (1/(1 - V² r²/(R² c²)) - 1/√(1 - V² r²/(R² c²))) r dr

Case 2:
2πρHc² ∫₀ᴿ (1/(1 - (V²/c²)(1 - r/R)²) - 1/√(1 - (V²/c²)(1 - r/R)²)) r dr

Case 1 for sphere:
4πρc² ∫₀ᴿ (1/(1 - V² r²/(c² R²)) - 1/√(1 - V² r²/(c² R²))) r² dr

Case 2 for sphere:
4πρc² ∫₀ᴿ (1/(1 - (V²/c²)(1 - r/R)²) - 1/√(1 - (V²/c²)(1 - r/R)²)) r² dr

Agree: @DontTalkDT (new version) @Dalesean027 (new version)
Disagree:
Neutral:

 

[Ghostimuscrime]

bro’s on an insane streak

 

[DontTalkDT]

Your formulas show up corrupted. Could you fix that and ideally show the derivation?

 

[Floxy178]

Your formulas show up corrupted. Could you fix that and ideally show the derivation?

Sure but I didn't understand how it shows up corrupted. Could you please say what parts need to be fixed? Because everything works for me.

 

[DontTalkDT]

Sure but I didn't understand how it shows up corrupted. Could you please say what parts need to be fixed? Because everything works for me.

Looks like this for me. These squares typically indicate that something isn't displayed properly.

 

[Ghostimuscrime]

Looks like this for me. These squares typically that something isn't displayed properly.

I think this is because you might be on mobile, desktop version shows me this

 

[Floxy178]

Looks like this for me. These squares typically that something isn't displayed properly.

Upper boundary of integral is R, as for other one, it's density. I can take screenshots and send them if you want. (idk how to replace that R to be honest)

 

[DontTalkDT]

I think this is because you might be on mobile, desktop version shows me this

Not on mobile, but now I know what it should look like.

But yeah, an explantion of the derivation would be good. How is the initial KE term in integral form reached?

 

[Floxy178]

Not on mobile, but now I know what it should look like.

But yeah, an explantion of the derivation would be good. How is the initial KE term in integral form reached?

So for them having same speed, I will look for relativistic KEs of rings r distance from center. Individual relativistic KE of a ring r distance from center is:
dKE = (γ(r) - 1)c² dm
Ring(or maybe calling it cylindrical shell would be better) with height of H, density of rho, thickness of dr (since it's infinitesmall change in r, what we're integrating over) will have:
dm = 2 * pi * r * rho * H * dr
Then we integrate it from r=0 to r=R for whole cylinder.

 

[DontTalkDT]

Ah, the physicist differential notation shenanigans. Always hated those. That's why I personally always started with the field definition of KE and then use integration by substitution to transform into polar coordinates. I think it looks ok in principle?

@Ugarik IIRC you also used that kind of approach to the problem. What do you think?


That aside, I'm not sure how this works in regard to relativistic density vs rest density. I have the suspicion that a correction might be needed here to account for the fact that we want a formula using rest mass, not the relativistic mass and perhaps length contraction. (Would time dilation matter as well? Uhhhh....)

 

[Floxy178]

That aside, I'm not sure how this works in regard to relativistic density vs rest density. I have the suspicion that a correction might be needed here to account for the fact that we want a formula using rest mass, not the relativistic mass and perhaps length contraction.

Oh if we're accounting for that then we can use proper(rest-mass) values and keep rest density. Though that'll depend on from what we're measuring, during motion or end result. In a case where we have the latter there'll be no need for accounting for length contraction.

Other that that we'll probably need replace R with proper radius (say R0), I guess it'll be R0 = ∫₀ᴿ γ(r) dr? (also R0/t instead of R/t for max speed, V)

Still I doubt that it's considered in such feats. Because in our case every part has its own speed, and therefore contraction will apply locally. So it'll be distorted rather than uniformly shorter, contraction will increase with r. (I'm not sure if it's visually noticeable or not, unfortunately I also couldn't find anything similar on internet)

I don’t think that these feats are intended to be bigger that what we actually observe tbh, overcomplicating it when its consideration in fiction itself is suspicious seems a little weird to me. I'm more in favor of going with lowball, unless it's really shown to be that way.

But if you think that it needs to be considered, I can change formulas according to that. (it won't be dramatical change anyway)

 

[DontTalkDT]

Quite frankly, I have no idea how exactly it works in calculation. I just figure that for instance a moving cube will have greater density than a stationary cube, even if at rest they are the same cube, as the cube is contracted in the direction of movement. So the local cloud density should probably not be constant due to the faster moving areas in the inertial frame end up with higher density if we assume that the clouds have uniform density prior to movement.

 

[Floxy178]

Quite frankly, I have no idea how exactly it works in calculation. I just figure that for instance a moving cube will have greater density than a stationary cube, even if at rest they are the same cube, as the cube is contracted in the direction of movement. So the local cloud density should probably not be constant due to the faster moving areas in the inertial frame end up with higher density if we assume that the clouds have uniform density prior to movement.

I mean if I can figure out this cube's rest properties then I can work with that, not what we observe. Alternatively we can go with density being higher as r approaches R but in this case I'll need to redo integrals.

If all parts had same speed, density would've become uniform, but here it applies locally so it completely changes solution of integral. If we intead change R and V to their rest values and keep density uniform, it'll result same as in the OP, we should just change values(R to R0, V=R/t to V=R0/t).

 

[Floxy178]

Bump

 

[Vzearr]

As far as I know, we don't have a method for calculating omnidirectional feats when speed is relativistic. I've seen it being calculated as being same fraction as classical KE formula (like calculating relativistic KE and dividing it by 2 or 6), but it isn't correct, as KE isn't quadratic with speed here, so distribution will be different from how those derived 1/4 or 1/12 treat it.

Difference becomes larger as speed is closer to c (method mentioned above gets higher results than actual), as for these omnidirectional feats, rise of KE isn't that big with speed approaching c compared to normal rel KE because most of its speed is significantly lower than c.

Yeah.

Math is fine. This is far too difficult to understand and took me an hour of reading to get it, so a calculator would be good.

I can make one I guess...? Might need some help.

 

[Floxy178]

Yeah.

Math is fine. This is far too difficult to understand and took me an hour of reading to get it, so a calculator would be good.

I can make one I guess...? Might need some help.

Would be very helpful, thanks

 

[KLOL506]

Quite frankly, I have no idea how exactly it works in calculation. I just figure that for instance a moving cube will have greater density than a stationary cube, even if at rest they are the same cube, as the cube is contracted in the direction of movement. So the local cloud density should probably not be constant due to the faster moving areas in the inertial frame end up with higher density if we assume that the clouds have uniform density prior to movement.

What do you suggest we should do?

 

[KLOL506]

I mean if I can figure out this cube's rest properties then I can work with that, not what we observe. Alternatively we can go with density being higher as r approaches R but in this case I'll need to redo integrals.

If all parts had same speed, density would've become uniform, but here it applies locally so it completely changes solution of integral. If we intead change R and V to their rest values and keep density uniform, it'll result same as in the OP, we should just change values(R to R0, V=R/t to V=R0/t).

@DontTalkDT

 

[Floxy178]

Bump

 

[LegendariumOfLies]

Bro is just him

 

[Floxy178]

Bump

 

[Floxy178]

Bump

 

[Nerd1435]

Yeah these are straight forward tbh, its just getting kinetic energy of differential element and integrating it over radius to get the net KE, it is the same as non-relativistic cloud dispersion in essence, the integration is slightly more complex that's all because of lorentz factor.

 

[Nerd1435]

Although honestly I think our standard for cloud dispersion trivializes it to a great extent by considering v(x)=(R-x)/t relation. The effect of imparted kinetic energy on change in speed as we move further along radius I don't think should be linearly approaching 0 should it? If we do assume the energy imparted projects mass constrained directionally to 2D (instead of a spherical wavefront), then can't we find the variation of speed?

Off the top of my head btw (not putting much thought into it nor did I work it out beforehand), it should be possible to work it out with linear momentum conservation. Assume clouds are inelastic.

Momentum P0 is imparted (assume) and this blast sweeps the cloud as a growing circular ring forms with radius r

Mass swept: M(r)= pi*r^2*rho*h, and this moves at v(r) speed, momentum is to be conserved. v(r)=P0/(pi*r^2*rho*h)
dr/dt=K/(r^2) just say K =P0/(pirhoh)
r^3/3=Kt or P0=pi*rho*h*r^3/(3*t)
r=(3Kt)^1/3
V=dr/dt=((3K)^1/3)*t^(-2/3)/3
And
KE=0.5*M(r)*V(r)^2 (since inelastic and they move together)
=0.5*r^2*(K/r^2)^2*pi*rho*h
=0.5*(P0^2/(pi*rho*h))*1/r^2
Gives (1/18)*pi*rho*h*r^4/t^2, we can just get it since we know it reaches circular radius r (Eg.10km) at time t (Eg.10sec)

 

[Floxy178]

Although honestly I think our standard for cloud dispersion trivializes it to a great extent by considering v(x)=(R-x)/t relation. The effect of imparted kinetic energy on change in speed as we move further along radius I don't think should be linearly approaching 0 should it? If we do assume the energy imparted projects mass constrained directionally to 2D (instead of a spherical wavefront), then can't we find the variation of speed?

Off the top of my head btw (not putting much thought into it nor did I work it out beforehand), it should be possible to work it out with linear momentum conservation. Assume clouds are inelastic.

Momentum P0 is imparted (assume) and this blast sweeps the cloud as a growing circular ring forms with radius r

Mass swept: M(r)= pi*r^2*rho*h, and this moves at v(r) speed, momentum is to be conserved. v(r)=P0/(pi*r^2*rho*h)
dr/dt=K/(r^2) just say K =P0/(pirhoh)
r^3/3=Kt or P0=pi*rho*h*r^3/(3*t)
r=(3Kt)^1/3
V=dr/dt=((3K)^1/3)*t^(-2/3)/3
And
KE=0.5*M(r)*V(r)^2 (since inelastic and they move together)
=0.5*r^2*(K/r^2)^2*pi*rho*h
=0.5*(P0^2/(pi*rho*h))*1/r^2
Gives (1/18)*pi*rho*h*r^4/t^2, we can just get it since we know it reaches circular radius r (Eg.10km) at time t (Eg.10sec)

Yeah if we assume dispersing being that way that should work mathematically. However we should wait for DT for now, if this gets accepted I can probably redo formulas in the OP.

But I personally disagree with this as it assumes that ring/blast goes outward with full concentrated swept mass while non reached clouds that are in range of r<=R remains static until that "boundary" reach them. So imo 1/12 one that finds how much minimum speed every part has to be for at time t all of clouds being outside of R is more reasonable as they move as a whole.

 

[Nerd1435]

Although honestly I think our standard for cloud dispersion trivializes it to a great extent by considering v(x)=(R-x)/t relation. The effect of imparted kinetic energy on change in speed as we move further along radius I don't think should be linearly approaching 0 should it? If we do assume the energy imparted projects mass constrained directionally to 2D (instead of a spherical wavefront), then can't we find the variation of speed?

Off the top of my head btw (not putting much thought into it nor did I work it out beforehand), it should be possible to work it out with linear momentum conservation. Assume clouds are inelastic.

Momentum P0 is imparted (assume) and this blast sweeps the cloud as a growing circular ring forms with radius r

Mass swept: M(r)= pi*r^2*rho*h, and this moves at v(r) speed, momentum is to be conserved. v(r)=P0/(pi*r^2*rho*h)
dr/dt=K/(r^2) just say K =P0/(pirhoh)
r^3/3=Kt or P0=pi*rho*h*r^3/(3*t)
r=(3Kt)^1/3
V=dr/dt=((3K)^1/3)*t^(-2/3)/3
And
KE=0.5*M(r)*V(r)^2 (since inelastic and they move together)
=0.5*r^2*(K/r^2)^2*pi*rho*h
=0.5*(P0^2/(pi*rho*h))*1/r^2
Gives (1/18)*pi*rho*h*r^4/t^2, we can just get it since we know it reaches circular radius r (Eg.10km) at time t (Eg.10sec)

Just realized, the inelastic collision assumption makes it such that K.E isn't conserved at all in this process, it only works with momentum conservation and only gives us the momentum. Most of it will have been wasted in heat and light. The kinetic energy I got here is just the rim's ke at this point

 

[Floxy178]

Just realized, the inelastic collision assumption makes it such that K.E isn't conserved at all in this process, it only works with momentum conservation and only gives us the momentum. Most of it will have been wasted in heat and light. The kinetic energy I got here is just the rim's ke at this point

Yeah it literally assumes maximum KE loss if I'm not mistaken.

 

[Nerd1435]

Yeah if we assume dispersing being that way that should work mathematically. However we should wait for DT for now, if this gets accepted I can probably redo formulas in the OP.

But I personally disagree with this as it assumes that ring/blast goes outward with full concentrated swept mass while non reached clouds that are in range of r<=R remains static until that "boundary" reach them. So imo 1/12 one that finds how much minimum speed every part has to be for at time t all of clouds being outside of R is more reasonable as they move as a whole.

By the time radius of gap reaches r=R, the initial 1/12 calc also achieves the same state as mine - the outer edge doesnt move at all and all mass concentrates on the rim.

Even in the one that I did, the linear mass density of the rim keeps increasing as it expands

 

[Nerd1435]

Yeah it literally assumes maximum KE loss if I'm not mistaken.

Yeah, well it is maximum KE loss by design, since I assumed inelasticity for the sake of simplicity, but yeah, it should be conserved energy. I should've figured beforehand since we literally assume gas particles undergo elastic collisions even during highschool-level kinetic gas theory lol

Edit: Actually since I'm not working with air but rather with clouds, it shouldn't be elastic

 

[Floxy178]

By the time radius of gap reaches r=R, the initial 1/12 calc also achieves the same state as mine - the outer edge doesnt move at all and all mass concentrates on the rim.

Even in the one that I did, the linear mass density of the rim keeps increasing as it expands

It being concentrated isn't problem alone, let me clarify.

For 1/12 case, we don't assume it being a concentrated ring at the end of motion in literal sense. Obviously edge part will also move, parts closer to edge too and they'll go beyond R, but the problem is, we don't know how much and can't realistically consider it for calculation from a single panel if you ask me. That just lets us claim that KE is > 1/12 m v^2.

But in your case, accounting for movement of clouds that aren't reached by rim, is against method, it doesn't make it lowballed version that serves as minimum value. Since it's an assumption that method needs to work in the first place, so it becomes unknown, not > 1/18 m v^2.

 

[Nerd1435]

It being concentrated isn't problem alone, let me clarify.

For 1/12 case, we don't assume it being a concentrated ring at the end of motion in literal sense. Obviously edge part will also move, parts closer to edge too and they'll go beyond R, but the problem is, we don't know how much and can't realistically consider it for calculation from a single panel if you ask me. That just lets us claim that KE is > 1/12 m v^2.

This comes back to whatever you said about what I've done: problem with the assumption.

Unless it is deliberately moved this way, there is no evidence that the velocity profile will even be remotely similar to v*(1-r/R) to say this will even be a low-end when someone just punches or screams and disperses clouds for example.

It needs to be derived. Instead, we can intuitively figure that the relation will not be linear. That's the whole reason I tried to get a relation using a more basic assumption instead.

But in your case, accounting for movement of clouds that aren't reached by rim, is against method, it doesn't make it lowballed version that serves as minimum value. Since it's an assumption that method needs to work in the first place, so it becomes unknown, not > 1/18 m v^2.

I did not account for movement of clouds not attached to the rim because I was working under the assumption of perfect inelasticity of clouds - the droplets just stick together and thus the density keeps increasing at the rim instead of the colloids bouncing off - was what I had in mind.

 

[Floxy178]

This comes back to whatever you said about what I've done: problem with the assumption.

Unless it is deliberately moved this way, there is no evidence that the velocity profile will even be remotely similar to v*(1-r/R) to say this will even be a low-end when someone just punches or screams and disperses clouds for example.

It needs to be derived. Instead, we can intuitively figure that the relation will not be linear. That's the whole reason I tried to get a relation using a more basic assumption instead.

I don't get how it's not a lowball. Obviously it's not linear normally but taking as linear gets minimum speed and thus KE for any part. What do you propose to consider movement beyond R?

I did not account for movement of clouds not attached to the rim because I was working under the assumption of perfect inelasticity of clouds - the droplets just stick together and thus the density keeps increasing at the rim instead of the colloids bouncing off - was what I had in mind.

Yeah I'm just making comparison to explain why accounting for it in 1/12 case doesn't contradict the method but just makes 1/12 value a lowball.

 

[Nerd1435]

I don't get how it's not a lowball. Obviously it's not linear normally but taking as linear gets minimum speed and thus KE for any part. What do you propose to consider movement beyond R?

It will be a lowball only if it is actually linear in reality - do we actually at least know or can we prove that this will lead to a lowball compared to reality? The relationship might even be something proportional to r^-x instead of 1-r/R, can this still be considered a lowball if that was the case?

Edit: r^-x is a bad example but ig you get what I'm trying to say

 

[Floxy178]

It will be a lowball only if it is actually linear in reality - do we actually at least know or can we prove that this will lead to a lowball compared to reality? The relationship might even be something proportional to r^-x instead of 1-r/R, can this still be considered a lowball if that was the case?

Edit: r^-x is a bad example but ig you get what I'm trying to say

Yeah we can because we're only taking distance traveled up to R into consideration and that gives us 1/12 m v^2. If you somehow could perfectly calculate with accordance of distances traveled beyond R within time t, that'd just increase the result. So any V(r) that we assume with linear is basically minimum speed/KE which is probably higher in reality if you consider what happens beyond R in t time.

 

[Nerd1435]

Yeah we can because we're only taking distance traveled up to R into consideration and that gives us 1/12 m v^2. If you somehow could perfectly calculate with accordance of distances traveled beyond R within time t, that'd just increase the result. So any V(r) that we assume with linear is basically minimum speed/KE which is probably higher in reality if you consider what happens beyond R in t time.

Sh** I really should sleep, don't mind whatever I said. A 6th grade kid should've realized this but I was in my own world completely overlooking the objective and that we already know it cleared out R in time t. So embarrassing.

 

[Floxy178]

No worries, take it easy

 

[Nerd1435]

No worries, take it easy

Hard to take it easy after messing up something an average 6th grade kid would've instantly realized... this is such a pattern man.. lack of focus and attention span overlooking really trivial things always messed up my performance, even costed me my chance at selection to my country's physics olympiad training camp. Anyway, my bad, none of whatever I said was necessary after the 1st comment.

 

[Floxy178]

Quite frankly, I have no idea how exactly it works in calculation. I just figure that for instance a moving cube will have greater density than a stationary cube, even if at rest they are the same cube, as the cube is contracted in the direction of movement. So the local cloud density should probably not be constant due to the faster moving areas in the inertial frame end up with higher density if we assume that the clouds have uniform density prior to movement.

I mean if I can figure out this cube's rest properties then I can work with that, not what we observe. Alternatively we can go with density being higher as r approaches R but in this case I'll need to redo integrals.

If all parts had same speed, density would've become uniform, but here it applies locally so it completely changes solution of integral. If we intead change R and V to their rest values and keep density uniform, it'll result same as in the OP, we should just change values(R to R0, V=R/t to V=R0/t).

Bump

Edit: on a second thought, I'm not sure if that'll work since I have no idea how to adjust t to that. So I'll go with density increasing which will add extra γ(r). For example case 1 will look like that:

2πϱHc² ∫₀ᴿ (γ(r) - 1) γ(r) r dr =

2πϱHc² ∫₀ᴿ (1/(1 - V²r²/(R²c²)) - 1/√(1 - V²r²/(R²c²))) r dr

 

[Nerd1435]

Bump

Edit: on a second thought, I'm not sure if that'll work since I have no idea how to adjust t to that. So I'll go with density increasing which will add extra γ(r). For example case 1 will look like that:

2πϱHc² ∫₀ᴿ (γ(r) - 1) γ(r) r dr =

2πϱHc² ∫₀ᴿ (1/(1 - V²r²/(R²c²)) - 1/√(1 - V²r²/(R²c²))) r dr

Why would you have an extra multiplication with the lorentz factor? Is it transformation on the differential element thickness? Or is it on the distance from center? I don't think it should apply in either of those cases

If you're observing from a rest frame, your requirement is for the differential element of the cloud to travel R-r as measure by you in time t as measured by you.

Isn't the "increasing density" already accounted for in the kinetic energy expression integral of (gamma-1)*c^2*dm? dm here should be invariant mass

 

[Floxy178]

DT asked for consideration of length contraction, that's why accounted for that.