# Relativistic formula for motion with constant acceleration

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• bobie
In summary, a comet F, located at a distance of 1 AU, is traveling towards a neutron star with a velocity of 0.99 c and an acceleration of 3 km/s^2. Assuming that the acceleration is uniform from the starting point to the star, it is estimated to reach the star after approximately 1000 seconds with a speed of 297*10^9 cm/s + 3*10^5*10^3. However, considering relativity, the final speed will be significantly smaller. The formula to find the actual final speed of the comet is unknown, but the formula ##at/\sqrt{1+a^2t^2/c^2}## may not be applicable in
bobie
Gold Member
Suppose a comet F ,at 1 AU distance, is traveling toward a massive body (a neutron star or other) with v= .99 c, suppose also that a = 3 km/s^2 and (to simplify calcs) that it is uniform from there to the star. It will hit the star after ca. 1000 seconds and its speed should equal C: 297*10^9 cm/s + 3*10^5*10^3

But if we take into account relativity, its speed will be much smaller, right? Now, what is the formula to find the actual final speed of F when it hits home?

I found this formula at/√1+ a^2 t^2/C^2 but it's useless here, and probably wrong altogether., since we can apply it to body B (with v0= 0) because it's not greatly affected by relativity, but the result is not exact.

Do you know the formula we must use to find the relativistic increase of velocity? from other calcs the velocity of F should be .9902 or so-
Thanks

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bobie said:
a proton star
Do you mean a neutron star?
bobie said:
a = 3 km/s
That's a velocity, not an acceleration. Did you mean 3000ms-2? Also, why is it accelerating? If you are thinking about gravity then you can't apply special relativity, and the proper acceleration is zero. You need to solve the geodesic equations in Schwarzschild spacetime (or Reissner-Nordstrom spacetime if you really meant protons). Edit: actually you can probably do it by conservation of energy if all you want is the impact speed. But you'd need the correct relativistic expressions for gravitational potential in your chosen metric and kinetic energy.

The expression ##at/\sqrt{1+a^2t^2/c^2}## is an exact expression for the velocity of a body undergoing constant proper acceleration in flat spacetime after coordinate time t in its initial rest frame. If I understand your scenario correctly it's completely irrelevant.

bobie
Ibix said:
Do you mean a neutron star?
. Edit: actually you can probably do it by conservation of energy if all you want is the impact speed. But you'd need the correct relativistic expressions for gravitational potential in your chosen metric and kinetic energy.

.
I corrected the typos, yes it' gravity (of a neutron star) and all I need is impact speed. If I understood correctly I can use PE the same as I do with body B, in that case Total energy would be 6,08881 EM0+3/4 = 6.82 M0 to which KE corresponds the velocity of .9918 C. Is that correct? Can you give me the right relativistic formula now?

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It's not really that simple because a lot of concepts don't translate across from Newtonian physics. However, you can make reasonable assumptions about what you mean. I think all the following is right, but it's possible I've made an algebraic slip somewhere.

A falling object like a comet is not accelerating in relativity. An object hovering at a fixed height is accelerating. That's the meaning of the equivalence principle - standing on a planet is indistinguishable from being in a rocket accelerating in space. The proper acceleration, ##a##, to hover at ##r_0## is ##a=c^2R_s/\left(2r_0^2\sqrt{1-R_s/r_0}\right)##. Strictly speaking, ##r_0## isn't the distance from the star, but it's not too far off. Substitute your acceleration and your distance and solve for the Schwarzschild radius, ##R_s##.

Next you need the Lorentz gamma factor, ##\gamma_0##, for the velocity of the infalling particle compared to a hovering observer at the start (you said v=0.99c at ##r_0##, and the Lorentz gamma factor is ##\gamma=1/\sqrt{1-v^2/c^2}##). Then you can calculate the gamma factor relative to a hovering observer at any other radius ##r## using$$\frac{\gamma}{\gamma_0}=\sqrt{\frac{r-R_s}{r_0-R_s}}$$Insert the radius, ##r##, of the neutron star and you should get an answer. As above this is slightly approximate since the Schwarzschild ##r## isn't quite naively interpretable as a radius, but it's not far off.

Note that I haven't checked that your acceleration numbers are reasonable. You may be better off working with a neutron star mass to calculate the Schwarzschild radius, then see what acceleration that gives at 1AU.

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bobie
Ibix said:
Paging @PeterDonis for quality assurance.

Ibix said:
Then you can calculate the gamma factor relative to a hovering observer at any other radius ##r## using
$$\frac{\gamma}{\gamma_0}=\sqrt{\frac{r-R_s}{r_0-R_s}}$$

How are you deriving this formula?

The way I would do this is by energy conservation: the increase in kinetic energy during the fall is equal to the decrease in potential energy, and the latter is just the difference ##GM / r## from the initial height to the radius of the neutron star. Note that this formula is exact in GR if ##r## is the radial coordinate, so you just need the radius of the neutron star (not its Schwarzschild radius but its actual radius), which I don't believe the OP has given, so we need that data point. Then just add the increase in kinetic energy to the original kinetic energy, and obtain the new relativistic gamma factor from that.

bobie said:
suppose also that a = 3 km/s^2 and (to simplify calcs) that it is uniform from there to the star

This is a drastically wrong assumption for a gravitating body over the range of heights you are using.

Ibix and bobie
PeterDonis said:
... Note that this formula is exact in GR if ##r## is the radial coordinate, so you just need the radius of the neutron star (not its Schwarzschild radius but its actual radius), which I don't believe the OP has given, so we need that data point. Then just add the increase in kinetic energy to the original kinetic energy, and obtain the new relativistic gamma factor from that.
This is a drastically wrong assumption for a gravitating body over the range of heights you are using.

Yes , I am aware, but the problem is already complex (before posting I had a Google search and I got half a dozen different formulae) and I thought it would be too difficult.

P.S could we derive PE considering that the exposure time is just 1/10?

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PeterDonis said:
How are you deriving this formula?
My argument was that for two four velocities ##U^\mu## and ##V^\nu##, ##\gamma=g_{\mu\nu}U^\mu V^\nu##. This is trivial in local Minkowski coordinates picked so that one of the four velocities represents rest.

If ##V^\nu## is the velocity of a hovering Schwarzschild observer then, in Schwarzschild coordinates, its only non-zero component is ##V^0=1/\sqrt{1-2GM/r}##, so the first expression reduces to ##\gamma=g_{00}U^0V^0## and we get ##U^0=\gamma/\sqrt{1-2GM/r}##.

Then I fed ##U^0=dt/d\tau## into the expression for energy at infinity, which Carroll gives as ##E=(1-2GM/r)dt/d\tau=\gamma\sqrt{1-2GM/r}##. That's conserved along a geodesic, so I can equate the right hand side at any r to any other (say ##r=R##), giving me an expression for ##\gamma(r)##.

Assuming my reasoning is correct, that gives $$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/R}{1-R_s/r}}$$...which does not reduce to the expression I gave due to me mucking up the algebra on the last line.

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bobie
PeterDonis said:
Note that this formula is exact in GR if rr is the radial coordinate, so you just need the radius of the neutron star (not its Schwarzschild radius but its actual radius), which I don't believe the OP has given, so we need that data point.
Indeed. I was assuming that the ##r## used in the final expression would be that of the surface of the star. The Schwarzschild radius is only needed as part of the maths.

bobie
Ibix said:
.

typical radius 10^6 but we need a star with R*100 so we get M*10^6 and from 10^12 cm to 10^11 we get roughly same PECan you work out the result with your formula? does it check out?

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PeterDonis said:
you just need the radius of the neutron star (not its Schwarzschild radius but its actual radius), which I don't believe the OP has given, so we need that data point. Then just add the increase in kinetic energy to the original kinetic energy, and obtain the new relativistic gamma factor from that.
If I need only the whole PE for F as for B, then it is really easy (and the actual parameters are not important.) Anyway , I worked out a rough model , can you check if it's OK?,

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bobie said:
Can you work out the result with your formula? does it check out?
You can do it yourself. Use my first formula in #4 to get whichever of ##a##, ##r_0##, or ##R_s## you don't know. Use the expression for the Lorentz ##\gamma## to write down your initial gamma factor. Then use the final expression in #7 to get the final ##\gamma## (using ##R## as your initial radius and ##r## as your impact radius, with ##\gamma_R## and ##\gamma_r## as the corresponding gamma factors). Then use the expression for the Lorentz gamma factor again to get the impact velocity.

bobie
Ibix said:
You can do it yourself. Use my first formula in #4 to get whichever of ##a##, ##r_0##, or ##R_s## you don't know. Use the expression for the Lorentz ##\gamma## to write down your initial gamma factor. Then use the final expression in #7 to get the final ##\gamma## (using ##R## as your initial radius and ##r## as your impact radius, with ##\gamma_R## and ##\gamma_r## as the corresponding gamma factors). Then use the expression for the Lorentz gamma factor again to get the impact velocity.
I thought the correct formula was in #7

bobie said:
I thought the correct formula was in #7
That's what I said.

Ibix said:
which does not reduce to the expression I gave due to me mucking up the algebra on the last line.

Yes, now what you've got looks correct to me.

bobie and Ibix
PeterDonis said:
Yes, now what you've got looks correct to me.
Thanks for checking.

PeterDonis said:
Yes, now what you've got looks correct to me.

Should I get ecaxctly the same result with both methods?

bobie said:
Should I get ecaxctly the same result with both methods?

Which methods?

bobie
PeterDonis said:
Which methods?

bobie said:
You haven't shown us what expression you are using for either the change in PE nor KE. And I don't know why (or even how) you'd add PE to my formula. If you show your working we might be able to comment. Please use LaTeX for the maths and definitions of symbols and explain what you are doing. See my post #7 for example. The formulae are important, but the words explaining why I chose those formulae are what tells the reader how I was thinking about them. They are what enable readers to understand how I got to my result and, if I'd done it explicitly in #4, would have enabled someone to point out that my final expression didn't follow from the derivation. I also define everything (or nearly everything) explicitly, except standard usages.

Diagrams can be useful, but a diagram almost literally covered in inconsistently typeset numbers with no explanation of where they come from is simply confusing.

bobie
Ibix said:
You haven't shown us what expression you are using for either the change in PE nor KE. And I don't know why (or even how) you'd add PE to my formula.
so you just need the radius of the neutron star (not its Schwarzschild radius but its actual radius), which I don't believe the OP has given, so we need that data point. Then just add the increase in kinetic energy to the original kinetic energy, and obtain the new relativistic gamma factor from that.

Body F has v = .99 c, so it has 7.09 M0 Mass 7.09-1 KE
I thought that Peter meant that I can add PE from 10^12 to 10^11 cm to get the total KE (6.83 E M0) and with reverse Lorentz figure out vinal velocity : .9918 c

Did I get it wrong?

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I cannot read your mind. It is your responsibility to communicate clearly ALL of what you are doing (edit: or at least answer my specific questions - posting your diagram for a fourth time did not do that, posting it for a fifth time won't either). If you cannot do that I cannot help you.

What formulae are you using for kinetic and potential energy? How are you combining them?

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Dale and bobie
Ibix said:

What formulae are you using for kinetic and potential energy? How are you combining them?
KE of an electron at .99 c is 6.9 (M-1) times the energy of an electron at rest (.511*106 eV)
PE is just GM/ hf - GM/h0. Right?

I reckoned that a mass of 10^36 g would give the value a mentioned in my primitive sketch and example, the numbers a very rough and round, you can use any othe example if you wish, what matters is the principle.
that is we find delta Pe, add it to Ke, check how many masses we get e reversing Lorentz we find final velocity.
Thanks

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bobie said:

Those are the same method. @Ibix' formula just calculates the ratio of gamma factors based on the difference in PE between the two heights. (@Ibix used the term "energy at infinity" in his post, but it amounts to the same thing.) My use of the term "add" was a bit misleading; it's actually best, as @Ibix's analysis shows, to think of the free-fall from starting to ending height as increasing the gamma factor by a ratio. Then you calculate the final velocity from the gamma factor.

bobie said:
PE is just GM/ hf - GM/h0. Right?

No. That is a Newtonian formula. This is a relativistic scenario.

bobie
Ibix said:
I cannot read your mind. It is your responsibility to communicate clearly ALL of what you are doing (edit: or at least answer my specific questions - posting your diagram for a fourth time did not do that, posting it for a fifth time won't either). If you cannot do that I cannot help you.

What formulae are you using for kinetic and potential energy? How are you combining them?

Hi Ibix ,
I mucked it up, I thought that I could add PE from A to B to the KE. I drew a better sketch:

I found KE by Lorentz, 6.08881 EMo, acceleration by GM/r^2, and PE by GM/ r, delta PE is rougly 1.32*10^9

can I just multiply PE by total Mass x 7.0888? Else what do I do? Is Schwartzild radius 3*10^8 cm?

Thanks a lot

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bobie said:
I found KE by Lorentz, 6.08881 EMo, acceleration by GM/r^2, and PE by GM/ r, delta PE is rougly 1.32*10^9

Ibix said:

#4 :A falling object like a comet is not accelerating in relativity. An object hovering at a fixed height is accelerating. That's the meaning of the equivalence principle - standing on a planet is indistinguishable from being in a rocket accelerating in space. The proper acceleration, ##a##, to hover at ##r_0## is ##a=c^2R_s/\left(2r_0^2\sqrt{1-R_s/r_0}\right)##. Strictly speaking, ##r_0## isn't the distance from the star, but it's not too far off. Substitute your acceleration and your distance and solve for the Schwarzschild radius, ##R_s##..
I am confused, why do I get r_sin that way? isn't it 2GM/c^2 = 3*10^8 cm? and what is r0 , the final impact at B?
and what is R or r in #7? if r is B then ##\sqrt( \frac{1-3*10^8/10^10 = .97}{1-3*10^8/ ??})##

bobie said:
I am confused, why do I get r_sin that way?
Because you specified the initial "acceleration due to gravity" of the comet to be a=3km/s2 at ##r_0##=1AU from the star. You didn't specify a mass (or Schwarzschild radius) for the star, so you need to calculate it. The concepts are different and the maths is more complex, but this is the relativistic equivalent of solving ##a=GM/r_0^2## for M.

If you prefer to specify a mass and an initial distance for the star you can use this formula to determine the "acceleration due to gravity" at that initial distance. You don't need it to proceed in this case.
bobie said:
isn't it 2GM/c^2 = 3*10^8 cm?
It is, but you didn't provide M. I provided a way to determine ##R_s## from the parameters you did provide.
bobie said:
and what is r0 , the final impact at B?
and what is R or r in #7?
I switched notations, it seems. In #4, ##r_0## is the initial distance from the star (or nearly so - it's the Schwarzschild r coordinate, which isn't far off). In #7, however, ##R## is the initial distance and ##r## is the final distance (the radius of the star). So ##R=r_0##. Apologies for the confusion.

Ibix said:
You don't need it to proceed in this case.
I didn't get that, anyway if I got the rest right, the formula is

##\sqrt( \frac{1-3*10^8/10^10 = .97}{1-3*10^8/ 10^12 = .9997})##

but the final velocity is less than the initial : .985 c

What is wrong, now?

bobie said:
What is wrong, now?
What is R? What does the symbol represent and what is its value?

What is r? What does the symbol represent and what is its value?

What is the formula you are trying to use? Have you put everything in in the correct place?

bobie
Ibix said:
What is R? What does the symbol represent and what is its value?
What is r? What does the symbol represent and what is its value?
What is the formula you are trying to use? Have you put everything in in the correct place?
R is A in my sketch =10^12 cm, r is B = 10^10
and the formula is the correct one in #7$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/R}{1-R_s/r}}$$.
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A}{1-R_s/B}}$$.

Agreed. That is not consistent with what you wrote in #28, though.

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bobie
Ibix said:
Agreed. That is not consistent with what you wrote in #28, though.
That's the equation in #28:
##\sqrt( \frac{1-3*10^8/10^10 = .97}{1-3*10^8/ 10^12 = .9997})##
and this is #30
##\sqrt( \frac{1- R_s:3*10^8/B:10^10 = .97}{1-3*10^8/ A:10^12 = .9997})##

and the result is .985

bobie said:
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A}{1-R_s/B}}$$
bobie said:
##\sqrt( \frac{1- R_s:3*10^8/B:10^10 = .97}{1-3*10^8/ A:10^12 = .9997})##

Ibix said:
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A =10^{12}}{1-R_s/B=10^{10}}}$$.
even switching values in the right places we get
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{.9997}{.97}}$$.

I get an impossible result , grater than c : 1.0152

Seriously? You can't even read back over the thread for the meaning of ##\gamma## without me holding your hand and tell you where to look?

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