Why does Newtonian dynamics break down at the speed of light

Click For Summary
Newtonian dynamics breaks down at the speed of light due to the fundamental assumptions about time, mass, and motion that Newton made, which do not hold true in relativistic contexts. As velocities approach the speed of light, Newton's laws become increasingly inaccurate because they are based on low-velocity approximations. The transition to special relativity introduces concepts like the Lorentz transformation, which alters the definitions of force and momentum, making them velocity-dependent. While Newton's laws can still be applied in a modified form using four-vectors in relativity, they are no longer the same as their classical definitions. Ultimately, the breakdown is due to the limitations of Newton's original framework rather than the laws themselves being entirely invalid.
nisarg
Messages
1
Reaction score
0
I tried searching the web for this topic but got an answer like "formulae used in classical mechanics are approximations or simplifications of more accurate formulae such as the ones in quantum mechanics and special relativity". My question is that why do the laws of Sir Isaac Newton no longer apply to objects at the speed of light? Is it the formulae that are causing the problem or the laws?

I really need a detailed explanation to understand this topic thoroughly, so if someone could help me on this, I would be more than grateful.

Thanks
 
Science news on Phys.org
Take a look at the Wikipedia article on General Relativity to start. There is a section that explains going from classical Newtonian mechanics to General Relativity.
 
nisarg said:
. My question is that why do the laws of Sir Isaac Newton no longer apply to objects at the speed of light? Is it the formulae that are causing the problem or the laws?

The laws are low velocity approximations. At higher and higher velocities, the laws get worse and worse.
 
nisarg said:
My question is that why do the laws of Sir Isaac Newton no longer apply to objects at the speed of light?

Do they? Galilean transformation fails at the speed of light, but Newton's laws of motion (in their original form) still apply. If it makes sense to use forces for photons is another question.
 
nisarg said:
My question is that why do the laws of Sir Isaac Newton no longer apply to objects at the speed of light? Is it the formulae that are causing the problem or the laws?
We just happen to live in a universe where Newtonian physics is not exact. It is perfectly possible to imagine a world where the laws are exact at all speeds (particle physics and some other fields would get problems , but let's ignore the microscopic part here), but experiments show we do not live in such a world.

DrStupid said:
but Newton's laws of motion (in their original form) still apply.
Acceleration is not parallel to force in general. How does that agree with Newtonian physics?
 
  • Like
Likes vanhees71
mfb said:
Acceleration is not parallel to force in general. How does that agree with Newtonian physics?

With replacement of Galilei transformation by Lorentz transformation Newton's "quantity of matter" becomes velocity dependent. In the result acceleration is no longer parallel to force.
 
I'm not too experienced with relativity so I'm not sure, but isn't the 4-force parallel to the 4-acceleration (unless the rest mass is changing)?

Chet
 
DrStupid said:
With replacement of Galilei transformation by Lorentz transformation Newton's "quantity of matter" becomes velocity dependent. In the result acceleration is no longer parallel to force.
A velocity-dependent scalar mass is not sufficient, you would need some sort of "vector mass". And I think that is beyond Newton's equation of motion. Even the Lorentz transformations on their own are beyond Newton's physics.
 
mfb said:
A velocity-dependent scalar mass is not sufficient, you would need some sort of "vector mass".

The velocity-dependent scalar mass

m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

results in

a = \left( {\frac{F}{{m_0 }} - v \cdot \frac{{v \cdot F}}{{m_0 \cdot c^2 }}} \right) \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}}

There is no need for some sort of "vector mass".

mfb said:
Even the Lorentz transformations on their own are beyond Newton's physics.

Of course it is. That's why I limited my statement to Newton's laws of motion.
 
  • Like
Likes harrylin
  • #10
DrStupid said:
a = \left( {\frac{F}{{m_0 }} - v \cdot \frac{{v \cdot F}}{{m_0 \cdot c^2 }}} \right) \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}}
Okay, if you add those extra terms - I would not call this "Newton's laws of motion" any more.
 
  • #11
mfb said:
Okay, if you add those extra terms - I would not call this "Newton's laws of motion" any more.

There are no extra terms.
 
  • #12
The 2nd law fails simply because it allows accelerating moving objects to increase their speed indefinitely, but as per the current valid theories and experimental results, nothing in our universe can move at a speed faster than c, when its speed is measured in an inertial ref. frame.
 
  • Like
Likes Puma
  • #13
DrStupid said:
There are no extra terms.
Compared to a=F/m?
 
  • #14
mfb said:
Compared to a=F/m?

Compared to F=dp/dt
 
  • #15
DrStupid said:
There is no need for some sort of "vector mass".

You've rolled it into the second term in the pre-factor.
 
  • #16
Vanadium 50 said:
You've rolled it into the second term in the pre-factor.

Which second term of which pre-factor?
 
  • #17
The part with the dot product. That makes it directional.
 
  • #18
Vanadium 50 said:
The part with the dot product. That makes it directional.

Are you confusing the equation for acceleration with the equation for quantity of matter (we better do not use the term mass at this place)? The latter does not contain such a part.
 
  • #19
As I said in post number 7, if expressed in terms of the 4-force and 4-acceleration, Newton's second law is recovered intact.

Chet
 
  • #20
nisarg said:
My question is that why do the laws of Sir Isaac Newton no longer apply to objects at the speed of light? Is it the formulae that are causing the problem or the laws?

Newton made a few assumptions about nature that turned out to be incorrect. For example, Newton's conception of time in the definitions given in the Principia as a quantity that moves forward independently without regard to motion (I'm paraphrasing here) was questioned later by Mach who influenced Einstein. Einstein also knew that Maxwell's equations predicted electromagnetic waves that all traveled with the same speed... but relative to what? After the rejection of the lumineferous ether largely due to the Michelson-Morley experiment Einstein proposed the two postulates of special relativity - one of which is that the speed of light is the same for all observers regardless of their state of motion. One of the consequences of this postulate (which does not suppose that time runs the same for everyone) is that the amount of time elapsed depends on an observers state of motion. This (and other consequences) of special relativity are only important when the speeds of objects approach the speed of light. If the speeds are low then the predictions made by SR reduce to Newtonian mechanics.

So... to answer the question. It is the axiomatic assumptions that are 'causing the problems'

As others have said, Newton's laws are still applicable in SR if you change the definition of force and momentum to be their four-vector definitions. However, in my limited experience with SR I've noticed that the concept of force (in the Newtonian sense) is not very convenient simply because of how messy this would get when applying the Lorentz transformations. The form of the laws look the same when using four-vectors (which is probably one of the reasons that four-momentum was defined the way it was!), but I would argue that this isn't really Newton's laws anymore.
 
  • #21
brainpushups said:
As others have said, Newton's laws are still applicable in SR if you change the definition of force and momentum to be their four-vector definitions. However, in my limited experience with SR I've noticed that the concept of force (in the Newtonian sense) is not very convenient simply because of how messy this would get when applying the Lorentz transformations. The form of the laws look the same when using four-vectors (which is probably one of the reasons that four-momentum was defined the way it was!), but I would argue that this isn't really Newton's laws anymore.
I used to feel the same way, thinking that the 4-force was just a contrived entity, designed specifically to recover Newton's 2nd law within the framework of SR. However, that feeling was dispelled when I saw the dazzling development in MTW in which they derived the equation for the 4-force acting on a stationary or moving charge within an electric and/or magnetic field and showing that it was described, independent of ma, by the magnitude of the charge times the contraction of the Faraday tensor with the 4-velocity vector. Did you not have the same response when you studied this?

Chet
 
  • Like
Likes brainpushups
  • #22
Cool. I have not seen that derivation. I'll have to check it out.
 
  • #23
Chestermiller said:
I used to feel the same way, thinking that the 4-force was just a contrived entity, designed specifically to recover Newton's 2nd law within the framework of SR. However, that feeling was dispelled when I saw the dazzling development in MTW in which they derived the equation for the 4-force acting on a stationary or moving charge within an electric and/or magnetic field and showing that it was described, independent of ma, by the magnitude of the charge times the contraction of the Faraday tensor with the 4-velocity vector. Did you not have the same response when you studied this?

Chet

I felt this way for a short time too. For me the four-force concept became clear when I started thinking about it as just a measure of the deviation of a particle's path from geodesic motion. Forces are defined that way too in Newtonian mechanics after all, as a measure of deviation from inertial motion.
 
  • #24
Cruz Martinez said:
I felt this way for a short time too. For me the four-force concept became clear when I started thinking about it as just a measure of the deviation of a particle's path from geodesic motion. Forces are defined that way too in Newtonian mechanics after all, as a measure of deviation from inertial motion.
That would apply more to acceleration than to force.

Chet
 
  • #25
Chestermiller said:
That would apply more to acceleration than to force.

Chet
yes, you're right, I will have to elaborate more on that thought.

If there's acceleration then we have force, but that alone can't account for the whole thing since different particles deviate from inertial motion differently when they interact with the same fields.
But still forces are a measure of how particles deviate from inertial motion,
I guess it is safe to say that force is a measure of how much a particle's motion is not inertial, once you take into account the dynamical parameters of the particle, what do you think?
 
Last edited:
  • #26
There's no stuff gets heavier at high speeds term in Newtons equations.
 
  • #27
DrStupid said:
Compared to F=dp/dt
Exacty - Newton's second law: "The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed" together with the definition: "The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly". Thus, in equation form : dp/dt = F

What turned out to be inexact in Newton's theory, at high speeds, were his assumptions about mass, time and length.
 
  • #28
harrylin said:
What turned out to be inexact in Newton's theory, at high speeds, were his assumptions about mass, time and length.

Which assumptions about mass do mean?
 
  • #29
It's true that dp/dt = F relativistically, where bold indicates 3-vectors. This can be derived from the more general dpμ/dτ=Fμ, which gets us into the 4-force that Chet was talking about. However, none of these terms is what it is in Newtonian mechanics.

Symon has a nice treatment of this in Chapter 14.
 
  • #30
Vanadium 50 said:
It's true that dp/dt = F relativistically, where bold indicates 3-vectors. This can be derived from the more general dpμ/dτ=Fμ, which gets us into the 4-force that Chet was talking about. However, none of these terms is what it is in Newtonian mechanics.

Symon has a nice treatment of this in Chapter 14.
Yes. For me, this was all I needed to be satisfied that the relationship was recovered intact in SR (even if the terms in component form were not the same). Seeing it satisfied in vector form made me very happy.

Chet
 

Similar threads

  • · Replies 48 ·
2
Replies
48
Views
4K
  • · Replies 130 ·
5
Replies
130
Views
14K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 27 ·
Replies
27
Views
5K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 139 ·
5
Replies
139
Views
17K