Relativistic formula for motion with constant acceleration

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
  • #1
bobie
Gold Member
720
2
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
grav.png


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
 

Attachments

  • grav.png
    grav.png
    5.2 KB · Views: 1,195
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes bobie
  • #3
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?
 
Last edited:
  • #4
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.

Edit: hmmmm - not totally sure about this. Paging @PeterDonis for quality assurance.
 
Last edited:
  • Like
Likes bobie
  • #5
Ibix said:
Paging @PeterDonis for quality assurance.

Yes? How may I help you? Please speak slowly and clearly into the microphone. :wink:

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.
 
  • Like
Likes Ibix and bobie
  • #6
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?
 
Last edited:
  • #7
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.o:)
 
Last edited:
  • Like
Likes bobie
  • #8
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.

Typical neutron star masses and radii should be easily googleable.
 
  • Like
Likes bobie
  • #9
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?
 
Last edited:
  • #10
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?,
 
Last edited:
  • #11
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.
 
  • Like
Likes bobie
  • #12
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
 
  • #13
bobie said:
I thought the correct formula was in #7
That's what I said.
 
  • #14
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.
 
  • Like
Likes bobie and Ibix
  • #15
PeterDonis said:
Yes, now what you've got looks correct to me.
Thanks for checking.
 
  • #16
PeterDonis said:
Yes, now what you've got looks correct to me.

Should I get ecaxctly the same result with both methods?
 
  • #17
bobie said:
Should I get ecaxctly the same result with both methods?

Which methods?
 
  • Like
Likes bobie
  • #18
PeterDonis said:
Which methods?
Adding PE and ibix' formula
 
  • #19
bobie said:
Adding PE and ibix' formula
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.
 
  • Like
Likes bobie
  • #20
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.
I don't wanto to add PE to your formula, Peter said
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?
 
Last edited:
  • #21
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?
 
Last edited:
  • Like
Likes Dale and bobie
  • #22
Ibix said:
I cannot read your mind. It is your responsibility to communicate clearly ALL of what you are doing. If you cannot do that I cannot help you.

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
 
Last edited:
  • #23
bobie said:
Adding PE and ibix' formula

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.
 
  • Like
Likes bobie
  • #24
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:

grav3.png


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
 

Attachments

  • grav3.png
    grav3.png
    9.4 KB · Views: 574
Last edited:
  • #25
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
You are using Newtonian formulae for gravity and potential energy, when you specifically started this thread asking about relativistic formulae. Please re-read post #11 and follow the instructions there, or tell me what you don't understand about them.
 
  • #26
Ibix said:
You are using Newtonian formulae for gravity and potential energy, when you specifically started this thread asking about relativistic formulae. Please re-read post #11 and follow the instructions there, or tell me what you don't understand about them.

#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/ ??})##
 
  • #27
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.
 
  • #28
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?
 
  • #29
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?
 
  • Like
Likes bobie
  • #30
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$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/R}{1-R_s/r}}$$.
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A}{1-R_s/B}}$$.
 
  • #31
Agreed. That is not consistent with what you wrote in #28, though.
 
Last edited:
  • Like
Likes bobie
  • #32
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
 
  • #33
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})##
Try reading them carefully.
 
  • #34
Ibix said:
Try reading them carefully.
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A =10^{12}}{1-R_s/B=10^{10}}}$$.
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
 
  • #35
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?
 
  • Like
Likes member 587159 and berkeman
<h2>1. What is the relativistic formula for motion with constant acceleration?</h2><p>The relativistic formula for motion with constant acceleration is given by <strong>x = x<sub>0</sub> + v<sub>0</sub>t + (1/2)at<sup>2</sup></strong>, where x is the final position, x<sub>0</sub> is the initial position, v<sub>0</sub> is the initial velocity, a is the constant acceleration, and t is the time elapsed.</p><h2>2. How does the relativistic formula differ from the classical formula for motion with constant acceleration?</h2><p>The relativistic formula takes into account the effects of special relativity, such as time dilation and length contraction, which are not accounted for in the classical formula. This is important at high speeds, where the effects of special relativity become significant.</p><h2>3. Can the relativistic formula be used for any type of motion?</h2><p>No, the relativistic formula is specifically for motion with constant acceleration. For other types of motion, such as non-constant acceleration or circular motion, different equations are needed.</p><h2>4. How is the relativistic formula derived?</h2><p>The relativistic formula is derived from the principles of special relativity, which state that the laws of physics should be the same for all observers in uniform motion. By applying these principles to the equations of motion, the relativistic formula can be derived.</p><h2>5. What are some real-life applications of the relativistic formula for motion with constant acceleration?</h2><p>The relativistic formula is used in many fields, including astrophysics, particle physics, and engineering. It is used to calculate the motion of objects at high speeds, such as spacecraft, particles in accelerators, and projectiles. It also helps in understanding the behavior of objects near the speed of light, which is important for technologies such as GPS and particle accelerators.</p>

1. What is the relativistic formula for motion with constant acceleration?

The relativistic formula for motion with constant acceleration is given by x = x0 + v0t + (1/2)at2, where x is the final position, x0 is the initial position, v0 is the initial velocity, a is the constant acceleration, and t is the time elapsed.

2. How does the relativistic formula differ from the classical formula for motion with constant acceleration?

The relativistic formula takes into account the effects of special relativity, such as time dilation and length contraction, which are not accounted for in the classical formula. This is important at high speeds, where the effects of special relativity become significant.

3. Can the relativistic formula be used for any type of motion?

No, the relativistic formula is specifically for motion with constant acceleration. For other types of motion, such as non-constant acceleration or circular motion, different equations are needed.

4. How is the relativistic formula derived?

The relativistic formula is derived from the principles of special relativity, which state that the laws of physics should be the same for all observers in uniform motion. By applying these principles to the equations of motion, the relativistic formula can be derived.

5. What are some real-life applications of the relativistic formula for motion with constant acceleration?

The relativistic formula is used in many fields, including astrophysics, particle physics, and engineering. It is used to calculate the motion of objects at high speeds, such as spacecraft, particles in accelerators, and projectiles. It also helps in understanding the behavior of objects near the speed of light, which is important for technologies such as GPS and particle accelerators.

Similar threads

  • Special and General Relativity
Replies
33
Views
2K
  • Special and General Relativity
2
Replies
55
Views
3K
  • Special and General Relativity
2
Replies
45
Views
3K
  • Special and General Relativity
Replies
34
Views
1K
  • Special and General Relativity
2
Replies
40
Views
2K
  • Special and General Relativity
Replies
29
Views
1K
  • Special and General Relativity
Replies
21
Views
1K
  • Special and General Relativity
2
Replies
36
Views
3K
Replies
13
Views
999
  • Special and General Relativity
3
Replies
75
Views
3K
Back
Top