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is λ=h/mv the same equation as λ=hc/E ? Watch

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    (Original post by 3nTr0pY)
    Always. When v = 0 you thus get p = mv, which is what you're used to.
    When v=0, you get p=0
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    (Original post by agostino981)
    When v=0, you get p=0
    Yeah that's right, but my point was that the form of p approaches mv as v approaches 0. So when v is small the form of the full equation will be very close to p = mv.

    Alternatively you can take the non-relativistic limit:

    \lim_{c\to \infty}\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = mv


    i.e. in normal Newtonian physics the speed of light is infinite and so p = mv.
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    (Original post by 3nTr0pY)
    Yeah that's right, but my point was that the form of p approaches mv as v approaches 0. So when v is small the form of the full equation will be very close to p = mv.

    Alternatively you can take the non-relativistic limit:

    \lim_{c\to \infty}\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = mv


    i.e. in normal Newtonian physics the speed of light is infinite and so p = mv.
    oook so what i understood so far is that λ=hc/E is for particles moving a relativistic speeds but you said it is because most of of the energy is due to its momentum i dont get that and a second issue from where did you get the eqn E=p/c? please help
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    (Original post by >>MMM<<)
    oook so what i understood so far is that λ=hc/E is for particles moving a relativistic speeds but you said it is because most of of the energy is due to its momentum i dont get that and a second issue from where did you get the eqn E=p/c? please help
    These are the big equations that explain pretty much all you need to know:

     E^2 = p^2c^2 + m^2c^4

    p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}

    When mass = 0, objects travel at the speed of light c. This gives 0/0 in the bottom equation so you can't use it. Instead you have to use the top equation. Putting mass = 0 gives:

     p = E/c

    However, if mass is not quite zero e.g. an electron, but still small, and p, the momentum, is very big relative to m, you get

     p \approx E/c

    But the full equation for the momentum of a particle with mass (independent of the energy term) is still the bottom equation. i.e.

    p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}
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    (Original post by 3nTr0pY)
    These are the big equations that explain pretty much all you need to know:

     E^2 = p^2c^2 + m^2c^4

    p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}

    When mass = 0, objects travel at the speed of light c. This gives 0/0 in the bottom equation so you can't use it. Instead you have to use the top equation. Putting mass = 0 gives:

     p = E/c

    However, if mass is not quite zero e.g. an electron, but still small, and p, the momentum, is very big relative to m, you get

     p \approx E/c

    But the full equation for the momentum of a particle with mass (independent of the energy term) is still the bottom equation. i.e.

    p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}
    why when the mass=0 particles move to the speed of light (im sorry bear with me this was poorly explained to me in class)? and when p=E/c is put into λ=h/mv we get λ=hc/E right? and thanks
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    (Original post by >>MMM<<)
    why when the mass=0 particles move to the speed of light (im sorry bear with me this was poorly explained to me in class)? and when p=E/c is put into λ=h/mv we get λ=hc/E right? and thanks
    The proper expression is λ=h/p. So when mass = 0 we have λ= h/(E/c) = hc/E

    λ=h/mv does not work in general. It only works when p = mv is a good approximation, which is in the case of slow moving particles which have mass.
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    (Original post by >>MMM<<)
    why when the mass=0 particles move to the speed of light (im sorry bear with me this was poorly explained to me in class)?
    The simplest explanation for photon is probably because photon is light so it must have light speed as it has been measured experimentally.

    If you seek a more technical explanation, General Relativity tells us that massless particle will follow a null geodesic.

    What is meant by geodesic is that imagine you have a flat paper, the shortest distance between two points will be a straight line. If you have a curved surface, then the line with shortest distance between two points on the surface may not be a straight line but a curved line, and this line is called a geodesic. So geodesic is the generalization of straight line to curved surface.

    In flat space time, we have something like

    d\tau^2=c^2 dt^2-d\vec{x}^2, and this is what defines a flat space-time, aka a Minkowski space or the metric of special relativity.

    In general relativity, this might gives you something more bizarre, but you can perform a coordinate transform such that your local space-time is flat.

    In layman's terms, d\tau is the infinitesimal interval between two points.

    And a null geodesic is that the length of the tangent of the geodesic is zero. This basically means d\tau=0.

    This gives us

    c^2 dt^2 - d\vec{x}^2=0

    c^2 dt^2 = d\vec{x}^2

    Hence

    c^2 = \frac{d\vec{x}^2}{dt^2}=\vec{v}^  2

    Thus we have the speed of a massless particle.

    |\vec{v}|=c

    *You may seek explanation why null geodesic defines the trajectory of light.

    You just literally do the same thing, but instead, because you know it is light, it must have the speed of light, hence you come to the same conclusion: When the infinitesimal interval as zero, the particle behaves like a photon, hence light must follow a null geodesic.
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    (Original post by agostino981)
    The simplest explanation for photon is probably because photon is light so it must have light speed as it has been measured experimentally.

    If you seek a more technical explanation, General Relativity tells us that massless particle will follow a null geodesic.

    What is meant by geodesic is that imagine you have a flat paper, the shortest distance between two points will be a straight line. If you have a curved surface, then the line with shortest distance between two points on the surface may not be a straight line but a curved line, and this line is called a geodesic. So geodesic is the generalization of straight line to curved surface.

    In flat space time, we have something like

    d\tau^2=c^2 dt^2-d\vec{x}^2, and this is what defines a flat space-time, aka a Minkowski space or the metric of special relativity.

    In general relativity, this might gives you something more bizarre, but you can perform a coordinate transform such that your local space-time is flat.

    In layman's terms, d\tau is the infinitesimal interval between two points.

    And a null geodesic is that the length of the tangent of the geodesic is zero. This basically means d\tau=0.

    This gives us

    c^2 dt^2 - d\vec{x}^2=0

    c^2 dt^2 = d\vec{x}^2

    Hence

    c^2 = \frac{d\vec{x}^2}{dt^2}=\vec{v}^  2

    Thus we have the speed of a massless particle.

    |\vec{v}|=c

    *You may seek explanation why null geodesic defines the trajectory of light.

    You just literally do the same thing, but instead, because you know it is light, it must have the speed of light, hence you come to the same conclusion: When the infinitesimal interval as zero, the particle behaves like a photon, hence light must follow a null geodesic.
    Nicely proved, but I guess the problem with that is that it's somewhat circular as you haven't proved the initial claim that massless particles follow null geodesics.

    Here's another way to show the same thing, although this also has the same problem as your proof:

    From the full equation we have:

     E^2 = p^2c^2 + m^2c^4

    with m = 0 gives

     E = pc

    and so

     \gamma mc^2 = \gamma mvc

    where  \gamma = 1/\sqrt(1-v^2/c^2)

    and so we find:

     v = c

    Don't shoot me for a lack of rigour but it conveys the basic idea perhaps more simply than introducing metric etc.
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    (Original post by 3nTr0pY)
    Here's another way to show the same thing, although this also has the same problem as your proof:

    From the full equation we have:

     E^2 = p^2c^2 + m^2c^4

    with m = 0 gives

     E = pc

    and so

     \gamma mc^2 = \gamma mvc

    where  \gamma = 1/\sqrt(1-v^2/c^2)

    and so we find:

     v = c

    Don't shoot me for a lack of rigour but it conveys the basic idea perhaps more simply than introducing metric etc.
    Just saying, when you have c=v, you get \gamma=\frac{1}{0} which means \frac{\gamma}{\gamma}=\frac{0}{0  } which is undefined.

    Apart from that, this is probably the simplest way to explain without mentioning relativity. Introducing limits \lim_{m\rightarrow 0} and \lim_{v\rightarrow c} maybe?
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    (Original post by agostino981)
    Just saying, when you have c=v, you get \gamma=\frac{1}{0} which means \frac{\gamma}{\gamma}=\frac{0}{0  } which is undefined.

    Apart from that, this is probably the simplest way to explain without mentioning relativity. Introducing limits \lim_{m\rightarrow 0} and \lim_{v\rightarrow c} maybe?
    Yeah, you're right, that would be best. Was in a bit of a rush though so decided instead to make every mathematician from East to West recoil in horror.
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    (Original post by 3nTr0pY)
    Yeah, you're right, that would be best. Was in a bit of a rush though so decided instead to make every mathematician from East to West recoil in horror.
    I kind of feel im lost here between your guys arguments/conversation can somebody just tell me a simple explanation why λ=hc/E which is used for photons applicable for particles moving near speed of light? (for A level) and thanks
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    (Original post by agostino981)
    Just saying, when you have c=v, you get \gamma=\frac{1}{0} which means \frac{\gamma}{\gamma}=\frac{0}{0  } which is undefined.

    Apart from that, this is probably the simplest way to explain without mentioning relativity. Introducing limits \lim_{m\rightarrow 0} and \lim_{v\rightarrow c} maybe?
    I kind of feel im lost here between your guys arguments/conversation can somebody just tell me a simple explanation why λ=hc/E which is used for photons applicable for particles moving near speed of light? (for A level) and thanks
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    (Original post by >>MMM<<)
    I kind of feel im lost here between your guys arguments/conversation can somebody just tell me a simple explanation why λ=hc/E which is used for photons applicable for particles moving near speed of light? (for A level) and thanks
    I've pretty much said all I can say really. Read over my posts again.

    It's your job to understand the explanations we've taken the time to give you, not ours.
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    (Original post by 3nTr0pY)
    I've pretty much said all I can say really. Read over my posts again.

    It's your job to understand the explanations we've taken the time to give you, not ours.
    -_- i didnt say i didnt appreciate your help i said i dont understand between your arguments which answer should I consider
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    really? a negative rating its not like I even offended anyone -_-
 
 
 
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