Understanding Resistance of Acceleration: Velocity & Force

[Jan Nebec]

Hello!

Why does resistance to acceleration depend on both the velocity of the object as well as the direction of the force?

In circular motion, we can measure the centripetal force and centripetal acceleration, then we can calculate objects mass. Speed remains constant.

But we would get the same result for force acting parallel to an object, where velocity would change?

Thanks!
Reference https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[Ibix]

Why does resistance to acceleration depend on both the velocity of the object as well as the direction of the force?

Fundamentally, the same reason as length contraction, time dilation, relativity of simultaneity and all other relativistic effects - we live in a universe whose spacetime is locally Minkowski.

You can derive it from the definition of momentum - and in that case the reason is the $\gamma$ that isn't present in Newtonian physics. Where does the $\gamma$ come from? Basically from the Minkowski nature of spacetime, via the Lorentz transforms.

Flip your question around: why shouldn't it depend on velocity? What answer would you consider acceptable to that question?

But we would get the same result for force acting parallel to an object, where velocity would change?

I'm not sure I follow your question. The formula for acceleration due to a force applied parallel to your motion is different to the one for a force perpendicular to your motion, yes, if that's what you are asking.

Note that this is all coordinate acceleration, so depends on choice of frame. Proper acceleration is a Lorentz scalar and has no such velocity or direction dependence.

 

[Chestermiller]

You do realize that an object traveling in circular motion is accelerating even if its speed is constant, correct?

 

[jbriggs444]

Why does resistance to acceleration depend on both the velocity of the object as well as the direction of the force?

One explanation is that force gives the rate of change of momentum over time: $F\ =\ \frac{dp}{dt}$

Under the rules of special relativity, momentum is given by: $p\ =\ m\gamma v$

The gotcha is that $\gamma$ is a function of velocity: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

If you have an acceleration perpendicular to velocity then gamma is a constant. If you have an acceleration parallel to the velocity, gamma changes with v and if you want to evaluate the first derivative of momentum then you have to invoke the product rule.

 

[russ_watters]

Why does resistance to acceleration depend on both the velocity of the object as well as the direction of the force?

Reference https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

I'm confused; you started a thread yesterday on why relativistic mass isn't used anymore and cited the Insight article on why. I think the answer is: it doesn't and you misread the article! Here's the passage you are probably referring to:

In addition, the resistance to acceleration depends on both the velocity of the object as well as the direction of the force and so relativistic mass cannot in general correspond to a generalisation of either.

This is true for Relativistic mass, and it's a problem, which is the main reason you were told yesterday for why it isn't used.

...unless I'm misinterpreting this (I am an engineer, not a physicist)...

 

[Jan Nebec]

I'm confused; you started a thread yesterday on why relativistic mass isn't used anymore and cited the Insight article on why. I think the answer is: it doesn't and you misread the article! Here's the passage you are probably referring to:

This is true for Relativistic mass, and it's a problem, which is the main reason you were told yesterday for why it isn't used.

...unless I'm misinterpreting this (I am an engineer, not a physicist)...

Yes, I was referring to that part of an article. But I needed more specific mathematical proof for such statement, since in high schools they don't teach acceleration from such deep approach.

 

[russ_watters]

Yes, I was referring to that part of an article. But I needed more specific mathematical proof for such statement, since in high schools they don't teach acceleration from such deep approach.

So what you are really asking for is a derivation or example problem for the equation showing the relativistic relation between the force acting on an object and its acceleration in the insight article?

 

[sweet springs]

But I needed more specific mathematical proof for such statement, since in high schools they don't teach acceleration from such deep approach.

Calculation of
\frac{d}{d\tau}\frac{\mathbf{v}}{\sqrt{1-\frac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}=\frac{d\mathbf{v}}{d\tau}\frac{d}{d\mathbf{v}}\frac{\mathbf{v}}{\sqrt{1-\frac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}
will be fine. Please try it.

 

[Jan Nebec]

Calculation of
\frac{d}{d\tau}\frac{\mathbf{v}}{\sqrt{1-\frac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}=\frac{d\mathbf{v}}{d\tau}\frac{d}{d\mathbf{v}}\frac{\mathbf{v}}{\sqrt{1-\frac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}
will be fine. Please try it.

Thank you! I think I've got what I needed :)