# B What will stop me from going at the speed of light?

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1. May 15, 2018

### Sas_TP123

Using Newton's 2nd law F=ma,
If you provide a constant force of 1mil newtons then an object will accelerate at 100m/s.
Using V = U + AT
I can say that (speed of light) 299,792,458 = 0 + 100T thus T = 2997924.58 seconds or I can achieve speed of light in 35 days or so.
Why is this not possible?or what's wrong with the math?

2. May 15, 2018

### Staff: Mentor

Have you read anything about Einstein's Special Relativity? The short answer is that that math just doesnt apply at high speed.

3. May 15, 2018

### Staff: Mentor

4. May 16, 2018

### olgerm

F=ma is not valid in special relativity.

5. May 16, 2018

### Staff: Mentor

Actually it is if F is the 4D force vector and a is the 4D acceleration vector.

6. May 17, 2018 at 10:49 AM

### VKint

I think the real question here is "why is Newton's second law invalid at high velocities?" Indeed, it seems arbitrary to declare that a rule like F = ma is valid at low velocities, but not when v becomes large relative to c. In fact, this seems to violate the principle of relativity itself, giving mathematically "preferential treatment" to objects travelling at certain velocities. However, this is not the case. In my opinion, the best way to understand why is to think carefully about which of the foundational concepts one learns about in mechanics (i.e., force, mass, energy, momentum, etc.) are the most "fundamental".

For example, in the Newtonian picture of mechanics, "force" is the fundamental quantity that determines the motion of all bodies relative to one another. However, even from a classical point of view, this is not necessarily the most convenient mathematical formulation. As is well-known to anyone who's taken kinematics, conservation laws frequently provide a vastly more efficient approach to solving problems than a direct application of Newton's second law.

Now, in the classical world, these conservation laws can be shown to follow from F = ma, and are thus "subsidiary" in some sense--in other words, you will never get the wrong answer by just applying F = ma (assuming you can successfully carry out the math). However, the utility of concepts like "energy" and "momentum" led physicists to wonder if perhaps these quantities are actually more "fundamental" than force. Classically, this line of thinking led to re-formulations of mechanics in terms of Lagrangians, Hamiltonians, and fields, each of which have their own "domains of utility" in which they're useful descriptions of nature.

All of the alternative schemes mentioned so far can be shown to be equivalent to Newton's original formulation of mechanics. However, when combined with a careful examination of Maxwell's equations for the electromagnetic field, taking a "momentum-first" approach to physics leads to a point of departure from Newton's second law (the three-dimensional version, that is). Maxwell's equations respect certain symmetries (called Lorentz transformations), and requiring the equations to apply unmodified in every inertial frame of reference implies that momentum transformations between frames must also respect these symmetries.

In other words, it's a question of where we "begin" mathematically--do we postulate that F = ma is always true and work from there, or do we take conservation of momentum, the electromagnetic field, and the principle of relativity to be fundamental and derive consequences from these? If we choose the former, then there's nothing preventing me from accelerating to the speed of light. If we take the latter approach, the speed of light naturally emerges as a strict upper bound on the relative speeds of any two observers.

Ultimately, the question can only be settled by experiment. At this point, the verdict is pretty much in.

Relativity wins.

7. May 17, 2018 at 11:59 AM

### phinds

You're missing the point. F=ma is a simplification of the actual formula, which is valid at ALL velocities. You're just getting hung up on the fact that for low speeds we can use the F=ma simplification. That is, you are actually ALWAYS using the full equation, you just automatically use the simplification F=ma at low velocities because that gives the right answer out to a large number of decimal places. You could use the full equation at low speeds but it would be a waste of time to do so.