How Does Zero Acceleration in One Frame Manifest as Nonzero in Another?

[GRDixon]

Author note: "_" means subscript, and "^" means superscript in the following. Any tips on how to write equations in PF posts will be appreciated.

The Lorentz transformation of acceleration components, from IRF K’ to K, can be accessed via Google on the Internet. The transformation of a_y^' contains two terms, say A+B. If it is stipulated that a_y^'=0, then A=-B, or

a_y=-(vu_y a_x/c^2)/(1-(vu_x)⁄c^2 ).

Clearly a_y can be nonzero in certain cases. Consequently we have the purely kinematic result that a_y^' may be zero, whereas a_y is nonzero. What are the implications of this curious result?

 

[starthaus]

Author note: "_" means subscript, and "^" means superscript in the following. Any tips on how to write equations in PF posts will be appreciated.

The Lorentz transformation of acceleration components, from IRF K’ to K, can be accessed via Google on the Internet. The transformation of a_y^' contains two terms, say A+B. If it is stipulated that a_y^'=0, then A=-B, or

a_y=-(vu_y a_x/c^2)/(1-(vu_x)⁄c^2 ).

Clearly a_y can be nonzero in certain cases. Consequently we have the purely kinematic result that a_y^' may be zero, whereas a_y is nonzero. What are the implications of this curious result?

You got the wrong transformation, the correct formula is much more complicated.

 

[yuiop]

Author note: "_" means subscript, and "^" means superscript in the following. Any tips on how to write equations in PF posts will be appreciated.

The Lorentz transformation of acceleration components, from IRF K’ to K, can be accessed via Google on the Internet. The transformation of a_y^' contains two terms, say A+B. If it is stipulated that a_y^'=0, then A=-B, or

a_y=-(vu_y a_x/c^2)/(1-(vu_x)⁄c^2 ).

Clearly a_y can be nonzero in certain cases. Consequently we have the purely kinematic result that a_y^' may be zero, whereas a_y is nonzero. What are the implications of this curious result?

In tex your equation would be:

$$a_y = -\frac{(v u_y a_x /c^2)}{(1- v u_x/ c^2 )}$$

(Just click the equation to see the code that produces it including the tex and and /tex tags in square barckets, or click reply and look at the tex code there).

I think it would help if you linked to a reference where you got the equation from. Do you have something in mind like equations (29), (35) and (36) of this reference: http://www.hep.princeton.edu/~mcdonald/examples/mechanics/matthews_ajp_73_45_05.pdf ?

 

[MikeLizzi]

Is this what you are looking for?

http://mysite.verizon.net/mikelizzi/Documents/Acceleration Transformations.htm

 

[GRDixon]

In tex your equation would be:

$$a_y = -\frac{(v u_y a_x /c^2)}{(1- v u_x/ c^2 )}$$

(Just click the equation to see the code that produces it including the tex and and /tex tags in square barckets, or click reply and look at the tex code there).

I think it would help if you linked to a reference where you got the equation from. Do you have something in mind like equations (29), (35) and (36) of this reference: http://www.hep.princeton.edu/~mcdonald/examples/mechanics/matthews_ajp_73_45_05.pdf ?

Many thanks for the link to the latex tutorial. Actually I used "Special Relativity", A.P.French, Eq. 5-25. But I like your reference (same transformations) better, since it can be linked to from the Internet. I did some algebra to get $$a_y '$$ in K terms. In your reference a zero $$a_y$$ in Eq. 35 would hold when the two terms to the right of the equal sign equal zero ... a condition which can be satisfied for nonzero values of $$a_y '$$. This doesn't seem to jibe with the fact that a zero $$F_y '$$ in frame K' implies a zero $$F_y$$ in frame K, according to the general force transformation.

 

[GRDixon]

Is this what you are looking for?

http://mysite.verizon.net/mikelizzi/Documents/Acceleration Transformations.htm

Actually I was using the more general version (Eq. 5_25 in the French reference). See my response to Kev above.

 

[starthaus]

Many thanks for the link to the latex tutorial. Actually I used "Special Relativity", A.P.French, Eq. 5-25. But I like your reference (same transformations) better, since it can be linked to from the Internet. I did some algebra to get $$a_y '$$ in K terms.

...yet, you got it wrong.

This doesn't seem to jibe with the fact that a zero $$F_y '$$ in frame K' implies a zero $$F_y$$ in frame K

Why would it? There is no reason.

according to the general force transformation.

It doesn't, this is not what the correct transformation shows. The general transform is:

$$\vec{F}=\frac{1}{1+(\vec{v}.\vec{u'})/c^2}[\vec{F'}/\gamma+\vec{v}/v^2((\vec{v}.\vec{F'})(1-1/\gamma)+(\vec{u'}.\vec{F'})v^2/c^2)]$$

where

$$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$$

$$\vec{v}$$ is the translational velocity between frames S and S'
$$\vec{u'}$$ is the particle velocity in frame S'
$$\vec{F}$$ is the force in frame S
$$\vec{F'}$$ is the force in frame S'

 

[MikeLizzi]

Excellent observarions all. I have concerned myself with transformations back and forth between the reference frame of the body that was accelerating, where u' = (0, 0, 0)

Now I see the unexpected (should have expected it!) behavior of transforming acceleration between reference frames when neither is the one in which the body is accelerating.

 

[yuiop]

Actually I was using the more general version (Eq. 5_25 in the French reference). See my response to Kev above.

The equations given by MikeLizzi are exactly the same as the ones given by the link I gave. Mike's first equation is the same as equation (29) in the link I gave and thrid equation is the same as equation (35). The transformation for acceleration in the y direction (36) is easily obtained fron the y transformation by changing all the y subscripts in Mike's third equation to z.

Now back to your original question. It does seem that the observation you made, that when the test particle if moving with constant velocity (zero acceleration) in the y' direction in the S' frame, it can appear to have non zero acceleration in the y direction in the S frame, is correct. This is related to the fact that the coordinate acceleration of a particle is not always in the same direction as the proper force and acceleration.

 

[GRDixon]

general transform is:

$$\vec{F}=\frac{1}{1+(\vec{v}.\vec{u'})/c^2}[\vec{F'}/\gamma+\vec{v}/v^2((\vec{v}.\vec{F'})(1-1/\gamma)+(\vec{u'}.\vec{F'})v^2/c^2)]$$

My reference ("Special Relativity" by A.P.French) provides the transformations for the components of force. According to that reference,

$$F_y = \frac{F_y ' / \gamma}{1 + v u_ ' /c^2}$$

(I hope I got the tex right.) According to that transformation, a zero y component of force in K maps to a zero y' component of force in K'. Yet, for selected values of particle velocity components, etc., the same is not true for the y (and y') components of acceleration. Although you're correct that F=ma doesn't apply, I nonetheless find it interesting that a measured, zero y (or z) component of acceleration in one frame could measure out to a nonzero measured value in the other frame.

 

[yuiop]

$$F_y = \frac{F_y ' / \gamma}{1 + v u_ ' /c^2}$$

(I hope I got the tex right.)

Nearly right. I think you meant to put an x or y subscipt after u_

P.S. Here is another link http://arxiv.org/PS_cache/physics/pdf/0507/0507099v1.pdf that might be helpful. See equations 23,25,26 for the transformations of acceleration and equations 41,44,45 for the transformations of force, where they use the notation $$\dot{x}=dx/dt$$ and $$\ddot{x} = d^2 x/dt^2$$.

Going by the link (Equation 44), you meant to say:

$$F_y = \frac{F_y ' / \gamma}{1 + v u_x ' /c^2}$$

and here is yet another link that gives the inverse transformation that you seem to be using: http://www.physics.princeton.edu/~mcdonald/examples/EM/jefimenko_ajp_64_618_96.pdf Does that seem to agree with equations in the A.P.French book?

 

[starthaus]

My reference ("Special Relativity" by A.P.French) provides the transformations for the components of force. According to that reference,

$$F_y = \frac{F_y ' / \gamma}{1 + v u_ ' /c^2}$$

This is not what you wrote in post 1, that was completely wrong. What you are writing now is just incomplete, I gave you the complete expression.

According to that transformation, a zero y component of force in K maps to a zero y' component of force in K'.

According to the complete expression, that is not true.

I nonetheless find it interesting that a measured, zero y (or z) component of acceleration in one frame could measure out to a nonzero measured value in the other frame.

SR is really interesting.

 

[GRDixon]

here is yet another link that gives the inverse transformation that you seem to be using: http://www.physics.princeton.edu/~mcdonald/examples/EM/jefimenko_ajp_64_618_96.pdf Does that seem to agree with equations in the A.P.French book?

Yes, Eq. 30 in your reference is the same as in the A.P.French book (which, unfortunately, I think may be out of print).

 

[yuiop]

My reference ("Special Relativity" by A.P.French) provides the transformations for the components of force. According to that reference,

$$F_y = \frac{F_y ' / \gamma}{1 + v u_ ' /c^2}$$
...

This is not what you wrote in post 1, that was completely wrong. What you are writing now is just incomplete, I gave you the complete expression...

GRDixon gave the equation for acceleration transformation in post 1 and the equation from force transformation here, so it is not surprising that they are not the same.

 

[yuiop]

My reference ("Special Relativity" by A.P.French) provides the transformations for the components of force. According to that reference,

$$F_y = \frac{F_y ' / \gamma}{1 + v u_ ' /c^2}$$

(I hope I got the tex right.) According to that transformation, a zero y component of force in K maps to a zero y' component of force in K'. Yet, for selected values of particle velocity components, etc., the same is not true for the y (and y') components of acceleration. ...

The references I have found seem to agree with your conclusion.

I think if Starthaus breaks his general force transformation equation into components, he will eventually arrive at the same conclusion.

 

[starthaus]

GRDixon gave the equation for acceleration transformation in post 1 and the equation from force transformation here, so it is not surprising that they are not the same.

Nope, his equation at post 1 is completely wrong, the correct one is:

$$a_y=a'_y(1+\frac{vu'_x}{c^2})^{-2}*\gamma^{-2}-u'_ya'_x\frac{v}{c^2}(1+\frac{vu'_x}{c^2})^{-3}*\gamma^{-2}$$

He's missing a power of -3 and he's missing the factor $$\gamma^{-2}$$. Other than that, everything is correct :-)

 

[yuiop]

Nope, his equation at post 1 is completely wrong, the correct one is:

$$a_y=a'_y(1+\frac{vu'_x}{c^2})^{-2}*\gamma^{-2}-u'_ya'_x\frac{v}{c^2}(1+\frac{vu'_x}{c^2})^{-3}*\gamma^{-2}$$

He's missing a power of -3 and he's missing the factor $$\gamma^{-2}$$. Other than that, everything is correct :-)

Yes, that is the correct acceleration transformation, and it can be seen that when $u'_y$ is not zero, a non-zero value for $a'_x$ results in a non zero value for $a_y$ even when $a'_y$ is zero.

 

[starthaus]

Yes, that is the correct acceleration transformation, and it can be seen that when $u'_y$ is not zero, a non-zero value for $a'_x$ results in a non zero value for $a_y$ even when $a'_y$ is zero.

Yes, such is relativity: strange. The point is that he got the formula wrong and you failed to notice iit repeatedly.

 

[yuiop]

Yes, such is relativity: strange. The point is that he got the formula wrong and you failed to notice iit repeatedly.

I was aware of it from the start and that is why we were referring to equations in linked references instead, because GRDixon was having trouble with tex formatting at the time and I was too lazy to post the correct equations since we had the linked references anyway.

 

[GRDixon]

Nope, his equation at post 1 is completely wrong, the correct one is:

$$a_y=a'_y(1+\frac{vu'_x}{c^2})^{-2}*\gamma^{-2}-u'_ya'_x\frac{v}{c^2}(1+\frac{vu'_x}{c^2})^{-3}*\gamma^{-2}$$

He's missing a power of -3 and he's missing the factor $$\gamma^{-2}$$. Other than that, everything is correct :-)

Ah! I finally see our disconnect. Your transformation from $$a_y$$ to $$a_y '$$ is correct. But if $$a_y$$ equals zero, then the two terms to the right of your equal sign must sum to zero. If you solve for $$a_y '$$, you'll see that a zero [text]a_y[/tex] can be manifest as a nonzero $$a_y '$$. That is what I found (and still find) so interesting.