O and P whose 4-velocity and 4-acceleration have a dot product of 0

[PhDeezNutz]

Homework Statement

An inertial observer $O$ has 4-velocity $\tilde{U}_0$ and a particle $P$ has a (variable) 4-acceleration $\tilde{A}$. If $\tilde{U}_0 \cdot \tilde{A} = 0$, what can you conclude about the speed of $P$ in the rest frame of $O$?

Relevant Equations

Two expressions derived in my class notes and in Rindler's book are

$\tilde{A} = \left( c \gamma \dot{\gamma}, \gamma^2 a_1 + \gamma \dot{\gamma} u_1, \gamma^2 a_2 + \gamma \dot{\gamma} u_2, \gamma^2 a_3 + \gamma \dot{\gamma} u_3 \right)$

$\dot{\gamma} = \frac{\gamma^3}{c^2} \left(\tilde{u} \cdot \tilde{a} \right)$

If $\tilde{U}_0 \cdot \tilde{A} = 0$ in one frame then I would imagine it is also zero in another frame because from my understanding is that dot products are invariant under boosts. So let's boost to the rest frame of O. In that frame

$\tilde{U}_{0T} = \left( c, 0,0,0 \right)$

and as stated in the relevant equations

$\tilde{A}_T = \left( c \gamma \dot{\gamma}, \gamma^2 a_1 + \gamma \dot{\gamma} u_1, \gamma^2 a_2 + \gamma \dot{\gamma} u_2, \gamma^2 a_3 + \gamma \dot{\gamma} u_3 \right)$

$\tilde{U}_{0T} \cdot \tilde{A}_T = 0 \Rightarrow c^2 \gamma \dot{\gamma} = 0 \Rightarrow \gamma \dot{\gamma} = 0$

Since $\gamma$ cannot be equal to zero we conclude $\dot{\gamma} = 0$

Using the second relevant equation

$\dot{\gamma} = \frac{\gamma^3}{c^2} \left(\tilde{u} \cdot \tilde{a} \right)$

$\tilde{u} \cdot \tilde{a} = 0$

I think this corresponds to particle P moving in uniform circular motion in the rest frame of O

 

[TSny]

From your result that $\dot \gamma = 0$, you should be able to answer the question about the speed of the particle from the point of view of observer $O$. The motion of the particle in $O$'s frame is not necessarily circular motion.

 

[PhDeezNutz]

From your result that $\dot \gamma = 0$, you should be able to answer the question about the speed of the particle from the point of view of observer $O$. The motion of the particle in $O$'s frame is not necessarily circular motion.

is P traveling with constant velocity in the frame of O?

Or at the very least constant speed?

 

[TSny]

is P traveling with constant velocity in the frame of O?

Or at the very least constant speed?

It's one of the above.

 

[PhDeezNutz]

It's one of the above.

constant speed. Right?

 

[TSny]

Express $\dot \gamma$ in terms of $u$ and $\dot u$, where $u$ is the speed of the particle. The condition $\dot \gamma = 0$ will then tell you something about either $u$ or $\dot u$.

 

[PeroK]

constant speed. Right?

What is $\gamma$ a function of?

 

[PhDeezNutz]

Express $\dot \gamma$ in terms of $u$ and $\dot u$, where $u$ is the speed of the particle. The condition $\dot \gamma = 0$ will then tell you something about either $u$ or $\dot u$.

I believe $\dot{\gamma} = \frac{\gamma^3}{c^2}\left(\tilde{u} \cdot \dot{\tilde{u}}\right)$

So $\tilde{u} \cdot \dot{\tilde{u}} = 0$

Meaning $\tilde{u}$ and $\dot{\tilde{u}}$ are orthogonal to each other. To me being orthogonal to each other doesn't mean either one is necessarily zero.

What is $\gamma$ a function of?

Speed. correct? I say speed instead of velocity because the $u^2$ will get us the same answer for +/- $u$

$\gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}$

 

[PeroK]

I believe $\dot{\gamma} = \frac{\gamma^3}{c^2}\left(\tilde{u} \cdot \dot{\tilde{u}}\right)$

So $\tilde{u} \cdot \dot{\tilde{u}} = 0$

Meaning $\tilde{u}$ and $\dot{\tilde{u}}$ are orthogonal to each other. To me being orthogonal to each other doesn't mean either one is necessarily zero.
Speed. correct? I say speed instead of velocity because the $u^2$ will get us the same answer for +/- $u$

$\gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}$

Technically, it is a function of $\vec u \cdot \vec u$.

 

[TSny]

I believe $\dot{\gamma} = \frac{\gamma^3}{c^2}\left(\tilde{u} \cdot \dot{\tilde{u}}\right)$

So $\tilde{u} \cdot \dot{\tilde{u}} = 0$

Meaning $\tilde{u}$ and $\dot{\tilde{u}}$ are orthogonal to each other. To me being orthogonal to each other doesn't mean either one is necessarily zero.

That's correct. But, $\tilde{u} \cdot \dot{\tilde{u}} = 0$ is a relation involving veclocity and acceleration vectors. See if you can show that $\tilde{u} \cdot \dot{\tilde{u}} = 0$ if and only if $u \dot u = 0$, where $u$ is speed (a scalar).

Hint: For any vector $\vec v$, $\,\,\, \vec v \cdot \vec v = |\vec v|^2$

 

[PeroK]

I believe $\dot{\gamma} = \frac{\gamma^3}{c^2}\left(\tilde{u} \cdot \dot{\tilde{u}}\right)$

So $\tilde{u} \cdot \dot{\tilde{u}} = 0$

Meaning $\tilde{u}$ and $\dot{\tilde{u}}$ are orthogonal to each other. To me being orthogonal to each other doesn't mean either one is necessarily zero.

The problem here is your logic. What you've done is something like: suppose $a = 0$, then $ab = 0$ then $a = 0$ or $b = 0$ then $a$ is not necessarily $0$.

 

[TSny]

Technically, it is a function of $\vec u \cdot \vec u$.

I don't see any difference between saying that $\gamma$ is a function of $\vec u \cdot \vec u$ and saying that $\gamma$ is a function of speed $u$.

$\vec u \cdot \vec u$ is identically equal to $u^2$. Maybe I'm overlooking something.

 

[PhDeezNutz]

That's correct. But, $\tilde{u} \cdot \dot{\tilde{u}} = 0$ is a relation involving veclocity and acceleration vectors. See if you can show that $\tilde{u} \cdot \dot{\tilde{u}} = 0$ if and only if $u \dot u = 0$, where $u$ is speed (a scalar).

Hint: For any vector $\vec v$, $\,\,\, \vec v \cdot \vec v = |\vec v|^2$

I'm having considerable difficulty.

I don't see how

$\tilde{u} \cdot \dot{\tilde{u}} = u_1 \dot{u}_1 + u_2 \dot{u}_2 + u_3 \dot{u}_3 = 0 \Rightarrow \sqrt{u_1^2 + u_2^2 + u_3^2} \sqrt{\dot{u}_1^2 + \dot{u}_3^2 + \dot{u}_3^2 } = 0$

If I "foil" out the terms underneath the radical I will get a bunch of cross terms so the two expressions are not equal. I'm drawing blank here.

 

[TSny]

I don't see how

$\tilde{u} \cdot \dot{\tilde{u}} = u_1 \dot{u}_1 + u_2 \dot{u}_2 + u_3 \dot{u}_3 = 0 \Rightarrow \sqrt{u_1^2 + u_2^2 + u_3^2} \color{green} {\sqrt{\dot{u}_1^2 + \dot{u}_2^2 + \dot{u}_3^2 }} = 0$

$\dot u \neq \sqrt{\dot{u}_1^2 + \dot{u}_2^2 + \dot{u}_3^2 }$

That is, $\frac{d}{dt}|\vec u| \neq |\frac{d}{dt}\vec u |$ .

What do you get if you take the time derivative of the identity $\vec u \cdot \vec u = u^2$?

 

[PeroK]

I'm having considerable difficulty.

I don't see how

$\tilde{u} \cdot \dot{\tilde{u}} = u_1 \dot{u}_1 + u_2 \dot{u}_2 + u_3 \dot{u}_3 = 0 \Rightarrow \sqrt{u_1^2 + u_2^2 + u_3^2} \sqrt{\dot{u}_1^2 + \dot{u}_3^2 + \dot{u}_3^2 } = 0$

If I "foil" out the terms underneath the radical I will get a bunch of cross terms so the two expressions are not equal. I'm drawing blank here.

Note that $\gamma$ is a real-valued function of a real variable $u^2$. If $u^2$ changes then $\gamma$ changes. That's it. You cannot change the speed without changing $\gamma$. You don't actually need to use calculus at all.

It doesn't matter about how clever you try to be with organising the acceleration to be orthogonal to the velocity or any of that stuff. If the speed changes, then $\gamma$ changes. QED

 

[PeroK]

I don't see any difference between saying that $\gamma$ is a function of $\vec u \cdot \vec u$ and saying that $\gamma$ is a function of speed $u$.

$\vec u \cdot \vec u$ is identically equal to $u^2$. Maybe I'm overlooking something.

Sorry, I didn't mean to cause confusion. Yes, it's the same thing.

 

[PhDeezNutz]

Note that $\gamma$ is a real-valued function of a real variable $u^2$. If $u^2$ changes then $\gamma$ changes. That's it. You cannot change the speed without changing $\gamma$. You don't actually need to use calculus at all.

It doesn't matter about how clever you try to be with organising the acceleration to be orthogonal to the velocity or any of that stuff. If the speed changes, then $\gamma$ changes. QED

So to answer the question “What can we say about P’s speed in O’s rest frame?”...it’s simply constant in O’s rest frame.

correct?

 

[PeroK]

So to answer the question “What can we say about P’s speed in O’s rest frame?”...it’s simply constant in O’s rest frame.

correct?

Yes.

 

[PhDeezNutz]

Can I bother you two gentlemen to take a look at this thread if and when you get the chance?

https://www.physicsforums.com/threads/superluminal-moving-points-4-velocity.988718/

It deals with superluminal 4-velocities (which I did not even know was a thing until recently). I was able to get the results for U and U^2 = -c^2 in the second post but I did so using what some would consider questionable mathematics (treating derivatives as quotients of differentials). Also I never thought a dot product of 4-vector with itself could be negative but apparently it can according to the problem statement in Rindler’s book.

How would one prove that U is tangent to the world line? I always just assumed derivatives were tangent to their respective curves, never have I ever thought to prove it.

 

[PhDeezNutz]

$\dot u \neq \sqrt{\dot{u}_1^2 + \dot{u}_2^2 + \dot{u}_3^2 }$

That is, $\frac{d}{dt}|\vec u| \neq |\frac{d}{dt}\vec u |$ .

What do you get if you take the time derivative of the identity $\vec u \cdot \vec u = u^2$?

You are extremely clever.

$\frac{d}{dt} \left( \tilde{u} \cdot \tilde{u} \right) = \frac{d}{dt} \left( u^2\right) \Rightarrow 2 \tilde{u} \dot{\tilde{u}} = 2 u \dot{u}$

We'v established that $\tilde{u} \cdot \dot{\tilde{u}} = 0$ so it must be that $u \dot{u} = 0$. So one or the other (or both have to be 0). If $u = 0$ (for all time) then it is trivially true $\dot{u} = 0$. A less restrictive case is when $u = constant$ and $\dot{u} = 0$.

So P's speed must be constant in O's rest frame at the very least (It could be 0 but that would be trivial).

 

[PhDeezNutz]

Correction

$2 \tilde{u} \cdot \dot{\tilde{u}} = 2 u \dot{u}$

 

[TSny]

You are extremely clever.

$\frac{d}{dt} \left( \tilde{u} \cdot \tilde{u} \right) = \frac{d}{dt} \left( u^2\right) \Rightarrow 2 \tilde{u} \dot{\tilde{u}} = 2 u \dot{u}$

I've seen this "trick" a number of times before.

We'v established that $\tilde{u} \cdot \dot{\tilde{u}} = 0$ so it must be that $u \dot{u} = 0$. So one or the other (or both have to be 0). If $u = 0$ (for all time) then it is trivially true $\dot{u} = 0$. A less restrictive case is when $u = constant$ and $\dot{u} = 0$.

So P's speed must be constant in O's rest frame at the very least (It could be 0 but that would be trivial).

Yes. The problem states that the 4-acceleration is variable. So, I think that rules out the trivial case of the particle staying at rest in O's frame.

Note that @PeroK 's argument is the most succinct. If $\dot \gamma = 0$ at all times then $\gamma$ must be a constant. By inspection of how $\gamma$ depends on $u$, you can then conclude without any calculation that $u$ is constant.

 

[PhDeezNutz]

I've seen this "trick" a number of times before.

Yes. The problem states that the 4-acceleration is variable. So, I think that rules out the trivial case of the particle staying at rest in O's frame.

Note that @PeroK 's argument is the most succinct. If $\dot \gamma = 0$ at all times then $\gamma$ must be a constant. By inspection of how $\gamma$ depends on $u$, you can then conclude without any calculation that $u$ is constant.

While that may be true, I feel like I benefited from both methods. @PeroK 's common sense one and your more rigorous one. It's always good to see the mathematics corroborate intuitive notions.

The question I have is: How does variable 4-acceleration (in some general frame) preclude the particle being at rest in O's rest frame?

My best attempt at an explanation would be as follows (some of which may be built on rocky assumptions):

If P has variable acceleration in one frame then it has variable acceleration in every other frame. (bad assumption? If not how do I prove this is true?). If that is the case the vector $\dot{\tilde{u}} \neq 0 \Rightarrow \tilde{u} \neq 0$. Therefore P is not at rest in O's rest frame. This may seem to be in conflict with $\dot{u} = 0$ but it is possible for the vector to change direction while maintaining the same speed.

 

[PeroK]

While that may be true, I feel like I benefited from both methods. @PeroK 's common sense one and your more rigorous one. It's always good to see the mathematics corroborate intuitive notions.

This is not right. I used an elementary property of any monotonic continuous function. There was nothing unmathematical about that. In this case, we have:
$$\gamma(t_1) \ne \gamma(t_2) \ \Leftrightarrow \ u^2(t_1) \ne u^2(t_2)$$
Which is at least as rigorous as taking a mathematical sledehammer to the problem!

 

[PhDeezNutz]

This is not right. I used an elementary property of any monotonic continuous function. There was nothing unmathematical about that. In this case, we have:
$$\gamma(t_1) \ne \gamma(t_2) \ \Leftrightarrow \ u^2(t_1) \ne u^2(t_2)$$
Which is at least as rigorous as taking a mathematical sledehammer to the problem!

At first glance it appeared to be a mere appeal to intuition but I now see that it’s not. You’ll have to forgive me because my knowledge of rigorous/proper mathematics is almost non-existent.