Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Prove that if \(x\) is time-like and \(x \cdot y=0,\) then \(y\) is space-like.

Short Answer

Expert verified
If \(x\) is time-like and \(x \cdot y = 0\), then \(y\) is space-like.

Step by step solution

01

Define Time-like and Space-like Vectors

A vector \(x\) is time-like if its squared magnitude is negative, i.e., \(x \cdot x < 0\). A vector \(y\) is space-like if its squared magnitude is positive, i.e., \(y \cdot y > 0\).
02

Use Orthogonality Property

Since we know \(x \cdot y = 0\), it means that vectors \(x\) and \(y\) are orthogonal to each other.
03

Analyze the Perpendicular Space

In a spacetime context, having a time-like vector \(x\), the space orthogonal to it will consist of vectors that are purely spatial in nature, assuming non-degenerate signatures. This implies the perpendicular space does not add any time component, suggesting \(y\) must be space-like.
04

Connect the Dots Mathematically

Given \(x\) is time-like, in a Minkowski space with signature \((+---)\), the time-like portion dominates, ensuring the only orthogonal solutions are space-like. Therefore, \(y\) falling in this constraint can only meet the condition \(y \cdot y > 0\), thus proving \(y\) is space-like.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Time-like vectors
In classical mechanics, spacetime vectors are essential for understanding various physical phenomena. A time-like vector is a crucial concept in this context. A vector is considered time-like when its squared magnitude is negative. In mathematical terms, if you have a vector \( x \), it is time-like if \( x \cdot x < 0 \). This condition ensures that the vector has a time component that is more influential than spatial dimensions.### Characteristics of Time-like Vectors
  • They have a predominant temporal part, which means an entity moving along such a vector experiences time more than space.
  • They are essential for describing motions influenced by gravity, as observed in general relativity.
  • Time-like vectors usually imply that an object's speed is less than the speed of light, making them relatable to observers.
Recognizing time-like vectors assists in deciding whether an entity is noticeably traversing through time compared to space.
Space-like vectors
Space-like vectors offer an intriguing perspective on the nature of motion through spacetime. A vector is termed space-like if its squared magnitude is positive.Mathematically, for a vector \( y \), it is space-like if \( y \cdot y > 0 \). This definition indicates that the vector primarily extends across spatial dimensions rather than time.### Features of Space-like Vectors
  • Space-like vectors suggest that the spatial component of motion dominates over time.
  • Motions following space-like vectors are those where the perceived speed exceeds the speed of light, thus they do not describe physical particles realistically.
  • In theoretical physics, space-like separations illustrate events that are not causally linked.
Understanding space-like vectors is pivotal for grasping how space and time interact, especially in the realms beyond our ordinary experiences.
Orthogonality in spacetime
Orthogonality is a familiar concept from mathematics, dealing with perpendicular vectors. In spacetime, orthogonality takes on a deeper meaning.In this framework, two vectors \( x \) and \( y \) are orthogonal if their dot product is zero: \( x \cdot y = 0 \). This condition implies that the vectors do not share common components, even in the dynamic spacetime structure.### Importance of Orthogonality in Spacetime
  • Orthogonality ensures that one vector, say time-like, has no impact on the other, which might be space-like.
  • In a relativistic sense, orthogonal vectors relate to the independence of different events or movements.
  • This concept supports the geometrical arrangement of vectors in spacetime, where specific orthogonal directions exhibit exclusive properties.
Grasping orthogonality in spacetime is crucial for understanding how vectors can relate without influencing each other across complex spatial and temporal dimensions.
Minkowski space
Minkowski space presents a foundational setting in the theory of relativity, reconciling time and three-dimensional space into a four-dimensional spacetime.This space is recognized by its signature \((+---)\), signifying one time dimension and three spatial dimensions. Here, distances behave differently under Lorentz transformations compared to Euclidean spaces. ### Characteristics of Minkowski Space
  • Enables the representation of relativistic physics where time and space are interwoven into a single entity.
  • It is a pseudo-Riemannian manifold, as it allows for both positive and negative distances, accommodating time-like, space-like, and null vectors.
  • Essential for resolving paradoxes in relativity, such as simultaneous events appearing different depending on the observer's state of motion.
Minkowski space is indispensable because it allows the blending of Einstein's relativity theories with classical mechanics, providing a broad framework to comprehend the cosmos.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two events that occur at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) and times \(t_{1}\) and \(t_{2} .\) Let \(\Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}\) and \(\Delta t=t_{2}-t_{1} .\) Write down the Lorentz transformation for \(\mathbf{r}_{1}\) and \(t_{1}\), and likewise for \(\mathbf{r}_{2}\) and \(t_{2},\) and deduce the transformation for \(\Delta \mathbf{r}\) and \(\Delta t\). Notice that differences \(\Delta \mathbf{r}\) and \(\Delta t\) transform in exactly the same way as \(\mathbf{r}\) and \(t\). This important property follows from the linearity of the Lorentz transformation.

Newton's first law can be stated: If an object is isolated (subject to no forces), then it moves with constant velocity. We know that this is invariant under the Galilean transformation. Prove that it is also invariant under the Lorentz transformation. [Assume that it is true in an inertial frame \(\mathcal{S}\), and use the relativistic velocity-addition formula to show that it is also true in any other \(\mathcal{S}^{\prime} .\) ]

Two particles \(a\) and \(b\) with masses \(m_{a}=0\) and \(m_{b}>0\) approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.]

The neutral pion \(\pi^{0}\) is an unstable particle (mass \(m=135\ \mathrm{MeV} / c^{2}\) ) that can decay into two photons, \(\pi^{0} \rightarrow \gamma+\gamma .\) (a) If the pion is at rest, what is the energy of each photon? (b) Suppose instead that the pion is traveling along the \(x\) axis and that the photons are observed also traveling along the \(x\) axis, one forward and one backward. If the first photon has three times the energy of the second, what was the pion's original speed \(v ?\)

A particle \(a\) traveling along the positive \(x\) axis of frame \(\mathcal{S}\) with speed 0.5c decays into two identical particles, \(a \rightarrow b+b,\) both of which continue to travel on the \(x\) axis. (a) Given that \(m_{a}=2.5 m_{b},\) find the speed of either \(b\) particle in the rest frame of particle \(a .\) (b) By making the necessary transformation on the result of part (a), find the velocities of the two \(b\) particles in the original frame S.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free