# Newbie - increase velocity = increase in mass ?

#### trigonatus

From a quick search of the forum is appears that an increase in velocity does not result in an increase in mass. If this is correct why can't objects be accelerated to light speed and beyond?

If every action causes an equal and opposite reaction, what force stops an object from reaching light speed?

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#### Nugatory

Mentor

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#### jtbell

Mentor
An object's energy increases without limit ($E \rightarrow \infty$) as $v \rightarrow c$.

As you give an object more and more energy, by continuing to "push" on it somehow, its speed continues to increase, but more and more "slowly" as it gets closer to c. It can never actually reach a speed of c, because that would require an infinite amount of energy, which is impossible.

#### trigonatus

Thank you both.

Why more and more slowly as it gets closer to c? Is there some type of additional resistance or counter force that causes the decrease in acceleration as c is approached?

#### Nugatory

Mentor
Thank you both.

Why more and more slowly as it gets closer to c? Is there some type of additional resistance or counter force that causes the decrease in acceleration as c is approached?
There is no counterforce needed - it follows from the way that speeds add.

You are almost certainly thinking in terms of the classical law of velocity addition, which says that if A is moving at speed $u$ relative to B, and C is moving at speed $v$ relative to B, then C is moving at speed $(u+v)$ relative to A. Using that law, a spaceship moving at .99c relative to A and at rest relative to B could fire its engines and accelerate to .02c relative to B - and then it would be moving at 1.01c relative to A.

But that's not really how speeds add - the classical $(u+v)$ is an approximation that only works for speeds that are small compared with the speed of light. The more accurate law that you have to use for speeds near that of light is $(u+v)/(1+uv)$; use this and you'll see that even though nothing prevents the ship from accelerating just as you would expect, it still can't get above c relative to any observer.

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#### trigonatus

Now, I understand.
Thank you.

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